삼각 불평등 목록

기하학에서 삼각형 불평등은 모든 삼각형 또는 특정 조건을 충족하는 모든 삼각형의 매개변수를 포함하는 불평등이다.불평등은 두 가지 다른 가치의 순서를 정해준다. 즉, "보다 작거나 같거나", "보다 크거나 같거나" 또는 "보다 크거나 같음"이다.삼각형 불평등에서 매개변수는 측면 길이, 반주기, 각도 측정, 해당 각도의 삼각함수 값, 삼각형의 면적, 측면의 중위수, 고도, 각 각도에서 반대쪽으로의 내부 각도 이등분선 길이, 측면의 수직 이등분선,임의 지점으로부터 다른 지점까지의 거리, 인라디우스, 엑스트라디우스, 곡예 및/또는 기타 수량.

달리 명시되지 않는 한 이 글은 유클리드 평면의 삼각형을 다루고 있다.

주 파라미터 및 표기법

삼각형 불평등에서 가장 일반적으로 나타나는 매개변수는 다음과 같다.

• 측면 길이 a, b 및 c;
• 반퍼미터 s = (a + b + c) / 2 (둘레 p의 절반);
• 각도는 (각도와 동일한 기호로 표시된 정점을 사용하여) a, b 및 c의 각도를 측정한다.
• 각도의 삼각함수 값
• 삼각형의 면적 T.
• 측면의 중위수 m_a, m_b 및 m_c(각각 측면의 중간점에서 반대 정점까지의 선 세그먼트의 길이)
• 고도 h_a, h_b 및 h_c(각각 한쪽에 수직이고 반대쪽 정점까지 도달하는 세그먼트의 길이)
• 내부 각도 이등분자 t_a, t_b 및 t의_c 길이(각각 정점에서 반대쪽으로 분할하고 정점의 각도를 이등분하는 세그먼트)
• 측면의 수직 이등분자 p_a, p_b 및 p_c(각각 중간점에서 한 면에 수직이고 다른 면 중 하나에 도달하는 세그먼트의 길이)
• 평면의 임의 지점 P에 끝점이 있는 선 세그먼트의 길이(예: P에서 정점 A까지의 세그먼트의 길이는 PA 또는 AP로 표시됨)
• 그 내접원의 반지름 r(, 동그라미는 삼각형에 새긴 반경, 접선하는 모든 3면),exradii ra, rb, 라머 반경(한 방접원. 접선은 다른 양측의 연장의 a, b, c는 접선 좌우의 각은 반경), 원의 외접원의 반경 R(반경은 삼각형은 지나가는 throug에 circumscribed.한 h꼭지점 3개).

측면 길이

기본적인 삼각불평등은

또는 동등하게

게다가.

오른쪽의 값이 가능한 가장 낮은 경계일 경우, 특정 등급의 삼각형이 퇴보된 영역 0에 접근함에 따라 점증적으로 접근한다.^{[1]: p. 259} 모든 긍정적인 a, b, c를 지탱하는 왼쪽 불평등은 네스빗의 불평등이다.

우리는 가지고 있다.

${\displaystyle 3\left({\frac {a}{b}+{b}{b}{b}}{b}}{b}}{b}+{\frac{b}}{b}}{prac{a}}}}{c}}}\frac {a}\right)+3+3.}$^{[2]: p.250, #82}

${\displaystyle abc\geq(a+b-c)(a-b+c)(-a+b+c)\cHB }$^{[1]: 페이지 260}

${\displaystyle {\frac {1}{3}\leq {a^{2}+b^{2}+c^{2}}:{(a+b+c)^{1}}<{\frac {1}{2}}.\cHB }$^{[1]: 페이지 261}

${\displaystyle{\sqrt{a+b-c}+{\sqrt{a-b+c}+{\sqrt{a}+{\sqrt{b}}}{\sqrt {c}}}.}$^{[1]: 페이지 261}

${\displaystyle a^{2}b(a-b)+b^{2}c(b-c)+c^{2}a(c-a)\geq 0.}$^{[1]: 페이지 261}

C 각도가 둔감한 경우(90°보다 큼)그때

${\displaystyle a^{2}+b^{2}}}<c^{2};}$

C가 급성인 경우(90° 미만)그때

${\displaystyle a^{2}+b^{2}}}>c^{2}.}$

C가 직각일 때 평등의 중간 경우는 피타고라스 정리다.

대체적으로.^{[2]: p.1, #74}

${\displaystyle a^{2}+b^{2}}}{\frac{c^{2}},{2}},}$

동등성은 이소체 삼각형의 꼭지점 각도가 180°에 근접할 때에만 한계에 접근한다.

삼각형의^{[3]: p. 153} 중심이 삼각형의 근골 안에 있으면,

${\displaystyle a^{2}<4bc,\b^{2}<4ac,\cHB c^{2}<4ab.}$

위의 모든 불평등은 a, b, c가 가장 긴 쪽이 둘레의 절반 이하인 기본 삼각 불평등을 따라야 하기 때문에 사실이지만, 다음의 관계는 모든 긍정적인 a, b, c에 대해 유지된다.^{[1]: p.267}

${\displaystyle {\frac {3}}{ab+bc+ca}\leq {\sqrt[{3}]{3}}\leq {\frac {a+b+c}{3}}},}$

a = b = c일 때만 균등하게 각 홀딩이것은 비등변수의 경우 변의 조화 평균이 기하 평균보다 작으며, 이는 다시 산술 평균보다 작다고 말한다.

각도

${\displaystyle \cos A+\cos B+\cos C\leq {\frac {3}{2}.}$ ^{[1]: 286 페이지}

$\displaystyle (1-\cos A)(1-\cos C)\geq \cos A\codot \cos C.}$^{[2]: p.21, #836}

${\displaystyle \cos ^{4}{\frac {A}{2}}+\cos ^{4}{\frac {B}{2}}+\cos ^{4}{\frac {C}{2}}}{\frac {s^{3}}}}}}}}}$

반정위 s의 경우, 등변위만 동일하다.^{[2]: p.13, #608}

${\displaystyle a+b+c\geq 2{\sqrt{bc}\cos A+2{\sqrt{ca}\cos B+2{\sqrt{ab}}\cos C.}$ ^{[4]: Thm.1}

${\displaystyle \sin A+\sin B+\sin C\leq {3{\sqrt{3}}{2}.}$ ^{[1]: 286 페이지}

${\displaystyle \sin ^{2}A+\sin ^{2}B+\sin ^{2}C\leq {\frac {9}{4}}.}$ ^{[1]: 286 페이지}

${\displaystyle \sin A\cdot \sin B\cdot \sin C\leq \left({\frac {\sin A+\sin B+\sin C}{3}}\right)^{3}\leq \left(\sin {\frac {A+B+C}{3}}\right)^{3}=\sin ^{3}\left({\frac {\pi }{3}}\right)={\frac {3{\sqrt {3}}}{8}}.}$ ^{[5]: 203 페이지}

$\displaystyle \sin A+\sin B\cdot \sin C\leq \varphi }$^{[2]: p.p.c., #3297}

여기서 $\phi {\displaystyle }$= ${\displaystyle 1\frac{\mathrm{}}{}}$+ 5 ${\displaystyle \frac{2}{\mathrm{}}}$, ${\displaystyle \varphi ={\frac {1+{\sqrt{5}}{$2}}: $황금비.$

${\displaystyle \sin {\frac {A}{2}}\cdot \sin {\frac {B}{2}}\c}\cdot \sin {\frac {C}{2}}\leq {\frac {1}{1}{8}}}}}.}$ ^{[1]: 286 페이지}

${\displaystyle \tan ^{2}{\frac {A}{2}}+\tan ^{2}{\frac {B}{2}}+\tan ^{2}{\frac {C}{2}}\geq 1.}$ ^{[1]: 286 페이지}

$\displaystyle \cot A+\cot B+\cot C\geq {\sqrt {3}.}$ ^[6]

${\displaystyle \sin A\cdot \cdot \cos B+\sin B\cdot \cos A\leq{3{\sqrt{3}}{4}}}.}$^{[2]: p.p.201, #309.2}

할라디우스 R과 인라디우스 r에 대해 우리는 가지고 있다.

${\displaystyle \max(\sin {\frac {A}{2}},\sin {\b}{B}{2}},\sin {\frac {\c}{2}}\오른쪽)\leq {\\frac {1}{1+{\sqr}{2r}}{R}}}}}}}}}}}}}}}}}$

삼각형이 60°^{[7]: Cor. 3} 보다 크거나 같은 정점 각도의 등각인 경우에만 동등하게,그리고

${\displaystyle \min \pin(\sin(\pin(\pin(\frac {A}{B}{2}}: 오른쪽)\sin(\pin) {\prac {1}{1}{1}{1-{\sqrt}{2r}}{R}\right}),}}}}$

삼각형이 60°^{[7]: Cor. 3} 보다 작거나 같은 정점 각도를 가진 등각인 경우에만 동등하게.

우리는 또한 가지고 있다.

${\displaystyle {\frac {r}{R}-{\sqrt{1-{\frac {2r}}}}\leq \cos A\leq {\leq{R}}}+{1-{\frac {2r}{R}}}}}}}}}}}}}}}}}}}}}$

또한 각 B, C도 마찬가지로, 삼각형이 이등변이고 정점각이 60° 이하인 경우 첫 번째 부분에서 동등하고, 정점각이 60° 이하인 경우만 두 번째 부분에서 동등하다.^{[7]: Prop. 5}

또한, a와 b의 반대편에 있는 임의의 두 각도 측정치 A와 B는 다음과^{[1]: p. 264} 같이 관련된다.

${\displaystyle A>B\quad {\text{if and only if}\quad a>b,}$

이등변 삼각형 정리 및 그 반대방향과 관련이 있으며, A = B인 경우에만 이등변 삼각형 정리라고 명시되어 있다.

유클리드 외부 각도 정리에 의해 삼각형의 외부 각도는 정반대 정점에 있는 내부 각도 중 하나보다 크다.^{[1]: p. 261}

${\displaystyle 180^{\circle }-A>\max(B,C).}$

점 D가 삼각형 ABC 내부에 있으면

$[\displaystyle \angle BDC]\angle A.}$^{[1]: 263 페이지}

급성^{[2]: p.26, #954} 삼각형은

${\displaystyle \cos ^{2}A+\cos ^{2}B+\cos ^{2}C<1,}$

역불평등이 둔부 삼각형을 유지하며

게다가, 사용하지 않는 삼각형의 경우, 우리는^{[8]: Corollary 3}

${\displaystyle {\frac {2R+r}{R}}{R}\leq {\sqrt{2}}\좌(\cos \좌)\cos \좌({\frac {A-C}{2}}\우)++}좌({\frac {B}{2}}\우)}$

하이포텐use AC가 있는 직각 삼각형일 경우에만 동등하게.

면적

웨이젠보크의 불평등은, 영역 T에 있어서는,^{[1]: p. 290}

${\displaystyle a^{2}+b^{2}+c^{2}}}\geq 4{\sqrt {3}\cdot T,}$

등변형 케이스에서만 동등하게이것은 하드와이거-핀슬러 불평등의 골격이다.

${\displaystyle a^{2}+b^{2}+c^{2}}\geq (a-b)^{2}+(b-c)^{2}+(c-a)^{2}+4{\sqrt{3}}\cdot T.}$

또,

${\displaystyle ab+bc+ca\geq 4{\sqrt {3}\cdot T}$^{[9]: 페이지 138}

그리고^{[2]: p.192, #340.3 [5]: p. 204}

${\displaystyle T\leq {\frac {abc}{2}}{\sqrt {\frac {a+b+c}{a^{3}+b^{3}+c^{3}+abc}}}\leq {\frac {1}{4}}{\sqrt[{6}]{\frac {3(a+b+c)^{3}(abc)^{4}}{a^{3}+b^{3}+c^{3}}}}\leq {\frac {\sqrt{3}}{4}{4}}{4}}()()^{2/3}.}$

T의 가장 오른쪽 상단에서 산술-기하계 평균 불평등을 사용하여 삼각형에 대한 등거리 불평등을 구한다.

${\displaystyle T\leq {\frac {\sqrt{3}}{36}}(a+b+c)^{2}={\frac {\sqrt{3}}{9}s^{2}}$ ^{[5]: 203 페이지}

반퍼미터 단위의이것은 때때로 주변 p의 관점에서 다음과 같이 명시된다.

${\displaystyle p^{2}\geq 12{\sqrt{3}}\cdot T,}$

등변 삼각형과 ^[10]동등하게이것은 에 의해 강화된다.

${\displaystyle T\leq {\frac {\sqrt{3}}{4}}(abc)^{2/3}.}$

보네센의 불평등은 또한 등측 불평등을 강화한다.

${\displaystyle \pi ^{2}(R-r)^{2}\leq(a+b+c)^{2}-4\pi T.}$

우리는 또한 가지고 있다.

${\displaystyle {\frac {9abc}{a+b+c}\geq 4{\sqrt{3}\cdot T}$ ^{[1]: 페이지 290[9]: 페이지 138}

등변형 케이스에서만 동등하게,

${\displaystyle 38T^{2}\leq 2s^{4}-a^{4}-a^{4}-b^{4}-c^{4}}}$^{[2]: p.p.207, #2807}

반퍼미터 단위 및

${\displaystyle {\frac {1}{a}+{\frac {1}{b}+{\frac {1}{c}}<{\frac {s}}{{\frac}}}{{\frac}}}{}}}{\frac}}}}{\frac}}T}}.}$^{[2]: p.88, #2188}

급성 삼각형에 대한 오노의 불평등(모든 각도가 90° 미만인 삼각형)이다

${\displaystyle 27(b^{2}+c^{2}-a^{2})^{2}(c^{2}+a^{2}-b^{2}^{2}(a^{2}+b^{2}-c^{2})^{2}}^{2}\leq (4T)^{6}}}}}}$

삼각형의 넓이는 근방의 넓이와 비교할 수 있다.

${\displaystyle {\frac{\text}{incircle}{\text{Area of triangle}}}\leq {\frac {\pi }{3{\sqrt{3}}}}}}}}$

등변 삼각형에 대해서만 동등하게.^[11]

내부 삼각형이 기준 삼각형의 둘레를 동일한 길이 세그먼트로 분할하도록 기준 삼각형에 새겨진 경우, 해당 영역의 비율은 다음과 같다^{[9]: p. 138} .

${\displaystyle {\frac{\text{newed triangle}{\text}{Reference triangle}}}\leq {\frac {1}{4}}.}$

A, B, C의 내부 각도 이등분자는 D, E, F에서 반대쪽을 만나도록 한다.그러면^{[2]: p.18, #762}

${\displaystyle {\frac{3abc}{4(a^{3}+b^{3}+c^{3}}}}}}}}}\leq {\frac {{\text{Area of triangle}\,DEF}{\text{Aream}}},ABC}\leq {\frac {1}{4}.}$

삼각형의 중앙선을 통과하는 선은 원래 삼각형의 면적 대비 작은 하위 영역의 비율이 최소한 4/9가 되도록 면적을 분할한다.^[12]

중위수와 중심

삼각형의$$ 세 중위수 $m{\displaystyle {}_{}}$ ${\displaystyle a_{}}$ m ${\displaystyle b_{}}$ m ${\displaystyle c_{}}$ ${\$는 각각 정점과 반대편의 중간점을 연결하며, 길이의 합은 만족한다^{[1]: p. 271} .

${\displaystyle {\frac{3}{4}(a+b+c)<m_{a}+m_{b}+m_{c}<a+b+c}}$

게다가^{[2]: p.12, #589}

${\displaystyle \left({\frac {m_{a}}{a}}}\{b}}\frac\{b}\오른쪽)^{2}+\frac\{m_{c}}}^{2}\geq {\frac}{9}}}}}},},}$

등변형 케이스에서만 동등하게, 그리고 인라디우스 r의 경우,^{[2]: p.22, #846}

${\displaystyle {\frac {m_{a}m_{b}m_{c}^{2}+m_{b}^{2}+m_{c}^{2}}\geq r.}$

만약_a 우리가 M , M_b , M , 그리고 M으로_c 원곡선과 교차점까지 확장된 중위들의 길이를 더 나타낸다면, 그러면^{[2]: p.16, #689} .

${\displaystyle {\frac {M_{a}}{m_{a}}+{\frac {M_{b}}}+{\frac {M_{c}}}{c_}}\geq 4.}$

중심 G는 중위수의 교차점이다.AG, BG, CG가 각각 U, V, W에서 원주를 만나도록 한다.그럼^{[2]: p.17#723} 둘 다

${\displaystyle GU+GV+GW\geq AG+BG+CG}$

그리고

${\displaystyle GU\cdot GV\cdot GW\geq AG\cdot BG\cdot CG;}$

게다가^{[2]: p.156, #S56}

${\displaystyle \sin GBC+\sin GCA+\sin GAB\leq {\frac {3}{2}.}$

급성^{[2]: p.26, #954} 삼각형은

${\displaystyle m_{a}^{2}+m_{b}^{2}+m_{c}^{2}>6R^{2}}$

in terms of the circumradius R, while the opposite inequality holds for an obtuse triangle.

Denoting as IA, IB, IC the distances of the incenter from the vertices, the following holds:^{[2]: p.192, #339.3}

${\displaystyle {\frac {IA^{2}}{m_{a}^{2}}}+{\frac {IB^{2}}{m_{b}^{2}}}+{\frac {IC^{2}}{m_{c}^{2}}}\leq {\frac {3}{4}}.}$

The three medians of any triangle can form the sides of another triangle:^{[13]: p. 592}

${\displaystyle m_{a}<m_{b}+m_{c},\quad m_{b}<m_{c}+m_{a},\quad m_{c}<m_{a}+m_{b}.}$

Furthermore,^{[14]: Coro. 6}

${\displaystyle \max\{bm_{c}+cm_{b},\quad cm_{a}+am_{c},\quad am_{b}+bm_{a}\}\leq {\frac {a^{2}+b^{2}+c^{2}}{\sqrt {3}}}.}$

Altitudes

The altitudes h_a , etc. each connect a vertex to the opposite side and are perpendicular to that side. They satisfy both^{[1]: p. 274}

${\displaystyle h_{a}+h_{b}+h_{c}\leq {\frac {\sqrt {3}}{2}}(a+b+c)}$

and

${\displaystyle h_{a}^{2}+h_{b}^{2}+h_{c}^{2}\leq {\frac {3}{4}}(a^{2}+b^{2}+c^{2}).}$

In addition, if ${\displaystyle a\geq b\geq c,}$ then^{[2]: 222, #67}

${\displaystyle a+h_{a}\geq b+h_{b}\geq c+h_{c}.}$

We also have^{[2]: p.140, #3150}

${\displaystyle {\frac {h_{a}^{2}}{(b^{2}+c^{2})}}\cdot {\frac {h_{b}^{2}}{(c^{2}+a^{2})}}\cdot {\frac {h_{c}^{2}}{(a^{2}+b^{2})}}\leq \left({\frac {3}{8}}\right)^{3}.}$

For internal angle bisectors t_a, t_b, t_c from vertices A, B, C and circumcenter R and incenter r, we have^{[2]: p.125, #3005}

${\displaystyle {\frac {h_{a}}{t_{a}}}+{\frac {h_{b}}{t_{b}}}+{\frac {h_{c}}{t_{c}}}\geq {\frac {R+4r}{R}}.}$

The reciprocals of the altitudes of any triangle can themselves form a triangle:^[15]

${\displaystyle {\frac {1}{h_{a}}}<{\frac {1}{h_{b}}}+{\frac {1}{h_{c}}},\quad {\frac {1}{h_{b}}}<{\frac {1}{h_{c}}}+{\frac {1}{h_{a}}},\quad {\frac {1}{h_{c}}}<{\frac {1}{h_{a}}}+{\frac {1}{h_{b}}}.}$

Internal angle bisectors and incenter

The internal angle bisectors are segments in the interior of the triangle reaching from one vertex to the opposite side and bisecting the vertex angle into two equal angles. The angle bisectors t_a etc. satisfy

${\displaystyle t_{a}+t_{b}+t_{c}\leq {\frac {3}{2}}(a+b+c)}$

in terms of the sides, and

${\displaystyle h_{a}\leq t_{a}\leq m_{a}}$

in terms of the altitudes and medians, and likewise for t_b and t_c .^{[1]: pp. 271–3} Further,^{[2]: p.224, #132}

${\displaystyle {\sqrt {m_{a}}}+{\sqrt {m_{b}}}+{\sqrt {m_{c}}}\geq {\sqrt {t_{a}}}+{\sqrt {t_{b}}}+{\sqrt {t_{c}}}}$

in terms of the medians, and^{[2]: p.125, #3005}

${\displaystyle {\frac {h_{a}}{t_{a}}}+{\frac {h_{b}}{t_{b}}}+{\frac {h_{c}}{t_{c}}}\geq 1+{\frac {4r}{R}}}$

in terms of the altitudes, inradius r and circumradius R.

Let T_a , T_b , and T_c be the lengths of the angle bisectors extended to the circumcircle. Then^{[2]: p.11, #535}

${\displaystyle T_{a}T_{b}T_{c}\geq {\frac {8{\sqrt {3}}}{9}}abc,}$

with equality only in the equilateral case, and^{[2]: p.14, #628}

${\displaystyle T_{a}+T_{b}+T_{c}\leq 5R+2r}$

for circumradius R and inradius r, again with equality only in the equilateral case. In addition,.^{[2]: p.20, #795}

${\displaystyle T_{a}+T_{b}+T_{c}\geq {\frac {4}{3}}(t_{a}+t_{b}+t_{c}).}$

For incenter I (the intersection of the internal angle bisectors),^{[2]: p.127, #3033}

${\displaystyle 6r\leq AI+BI+CI\leq {\sqrt {12(R^{2}-Rr+r^{2})}}.}$

For midpoints L, M, N of the sides,^{[2]: p.152, #J53}

${\displaystyle IL^{2}+IM^{2}+IN^{2}\geq r(R+r).}$

For incenter I, centroid G, circumcenter O, nine-point center N, and orthocenter H, we have for non-equilateral triangles the distance inequalities^{[16]: p.232}

${\displaystyle IG<HG,}$

${\displaystyle IH<HG,}$

${\displaystyle IG<IO,}$

and

${\displaystyle IN<{\frac {1}{2}}IO;}$

and we have the angle inequality^{[16]: p.233}

${\displaystyle \angle IOH<{\frac {\pi }{6}}.}$

In addition,^{[16]: p.233, Lemma 3}

${\displaystyle IG<{\frac {1}{3}}v,}$

where v is the longest median.

Three triangles with vertex at the incenter, OIH, GIH, and OGI, are obtuse:^{[16]: p.232}

${\displaystyle \angle OIH}$ > ${\displaystyle \angle GIH}$ > 90° , ${\displaystyle \angle OGI}$ > 90°.

Since these triangles have the indicated obtuse angles, we have

${\displaystyle OI^{2}+IH^{2}<OH^{2},\quad GI^{2}+IH^{2}<GH^{2},\quad OG^{2}+GI^{2}<OI^{2},}$

and in fact the second of these is equivalent to a result stronger than the first, shown by Euler:^[17][18]

${\displaystyle OI^{2}<OH^{2}-2\cdot IH^{2}<2\cdot OI^{2}.}$

The larger of two angles of a triangle has the shorter internal angle bisector:^{[19]: p.72, #114}

${\displaystyle {\text{If}}\quad A>B\quad {\text{then}}\quad t_{a}<t_{b}.}$

Perpendicular bisectors of sides

These inequalities deal with the lengths p_a etc. of the triangle-interior portions of the perpendicular bisectors of sides of the triangle. Denoting the sides so that ${\displaystyle a\geq b\geq c,}$ we have^[20]

${\displaystyle p_{a}\geq p_{b}}$

and

${\displaystyle p_{c}\geq p_{b}.}$

Segments from an arbitrary point

Interior point

Consider any point P in the interior of the triangle, with the triangle's vertices denoted A, B, and C and with the lengths of line segments denoted PA etc. We have^{[1]: pp. 275–7}

${\displaystyle 2(PA+PB+PC)>AB+BC+CA>PA+PB+PC,}$

and more strongly than the second of these inequalities is:^{[1]: p. 278} If ${\displaystyle AB}$ is the shortest side of the triangle, then

${\displaystyle PA+PB+PC\leq AC+BC.}$

We also have Ptolemy's inequality^{[2]: p.19, #770}

${\displaystyle PA\cdot BC+PB\cdot CA>PC\cdot AB}$

for interior point P and likewise for cyclic permutations of the vertices.

If we draw perpendiculars from interior point P to the sides of the triangle, intersecting the sides at D, E, and F, we have^{[1]: p. 278}

${\displaystyle PA\cdot PB\cdot PC\geq (PD+PE)(PE+PF)(PF+PD).}$

Further, the Erdős–Mordell inequality states that^{[21] [22]}

${\displaystyle {\frac {PA+PB+PC}{PD+PE+PF}}\geq 2}$

with equality in the equilateral case. More strongly, Barrow's inequality states that if the interior bisectors of the angles at interior point P (namely, of ∠APB, ∠BPC, and ∠CPA) intersect the triangle's sides at U, V, and W, then^[23]

${\displaystyle {\frac {PA+PB+PC}{PU+PV+PW}}\geq 2.}$

Also stronger than the Erdős–Mordell inequality is the following:^[24] Let D, E, F be the orthogonal projections of P onto BC, CA, AB respectively, and H, K, L be the orthogonal projections of P onto the tangents to the triangle's circumcircle at A, B, C respectively. Then

${\displaystyle PH+PK+PL\geq 2(PD+PE+PF).}$

With orthogonal projections H, K, L from P onto the tangents to the triangle's circumcircle at A, B, C respectively, we have^[25]

${\displaystyle {\frac {PH}{a^{2}}}+{\frac {PK}{b^{2}}}+{\frac {PL}{c^{2}}}\geq {\frac {1}{R}}}$

where R is the circumradius.

Again with distances PD, PE, PF of the interior point P from the sides we have these three inequalities:^{[2]: p.29, #1045}

${\displaystyle {\frac {PA^{2}}{PE\cdot PF}}+{\frac {PB^{2}}{PF\cdot PD}}+{\frac {PC^{2}}{PD\cdot PE}}\geq 12;}$

${\displaystyle {\frac {PA}{\sqrt {PE\cdot PF}}}+{\frac {PB}{\sqrt {PF\cdot PD}}}+{\frac {PC}{\sqrt {PD\cdot PE}}}\geq 6;}$

${\displaystyle {\frac {PA}{PE+PF}}+{\frac {PB}{PF+PD}}+{\frac {PC}{PD+PE}}\geq 3.}$

For interior point P with distances PA, PB, PC from the vertices and with triangle area T,^{[2]: p.37, #1159}

${\displaystyle (b+c)PA+(c+a)PB+(a+b)PC\geq 8T}$

and^{[2]: p.26, #965}

${\displaystyle {\frac {PA}{a}}+{\frac {PB}{b}}+{\frac {PC}{c}}\geq {\sqrt {3}}.}$

For an interior point P, centroid G, midpoints L, M, N of the sides, and semiperimeter s,^{[2]: p.140, #3164 [2]: p.130, #3052}

${\displaystyle 2(PL+PM+PN)\leq 3PG+PA+PB+PC\leq s+2(PL+PM+PN).}$

Moreover, for positive numbers k₁, k₂, k₃, and t with t less than or equal to 1:^{[26]: Thm.1}

${\displaystyle k_{1}\cdot (PA)^{t}+k_{2}\cdot (PB)^{t}+k_{3}\cdot (PC)^{t}\geq 2^{t}{\sqrt {k_{1}k_{2}k_{3}}}\left({\frac {(PD)^{t}}{\sqrt {k_{1}}}}+{\frac {(PE)^{t}}{\sqrt {k_{2}}}}+{\frac {(PF)^{t}}{\sqrt {k_{3}}}}\right),}$

while for t > 1 we have^{[26]: Thm.2}

${\displaystyle k_{1}\cdot (PA)^{t}+k_{2}\cdot (PB)^{t}+k_{3}\cdot (PC)^{t}\geq 2{\sqrt {k_{1}k_{2}k_{3}}}\left({\frac {(PD)^{t}}{\sqrt {k_{1}}}}+{\frac {(PE)^{t}}{\sqrt {k_{2}}}}+{\frac {(PF)^{t}}{\sqrt {k_{3}}}}\right).}$

Interior or exterior point

There are various inequalities for an arbitrary interior or exterior point in the plane in terms of the radius r of the triangle's inscribed circle. For example,^{[27]: p. 109}

${\displaystyle PA+PB+PC\geq 6r.}$

Others include:^{[28]: pp. 180–1}

${\displaystyle PA^{3}+PB^{3}+PC^{3}+k\cdot (PA\cdot PB\cdot PC)\geq 8(k+3)r^{3}}$

for k = 0, 1, ..., 6;

${\displaystyle PA^{2}+PB^{2}+PC^{2}+(PA\cdot PB\cdot PC)^{2/3}\geq 16r^{2};}$

${\displaystyle PA^{2}+PB^{2}+PC^{2}+2(PA\cdot PB\cdot PC)^{2/3}\geq 20r^{2};}$

and

${\displaystyle PA^{4}+PB^{4}+PC^{4}+k(PA\cdot PB\cdot PC)^{4/3}\geq 16(k+3)r^{4}}$

for k = 0, 1, ..., 9.

Furthermore, for circumradius R,

${\displaystyle (PA\cdot PB)^{3/2}+(PB\cdot PC)^{3/2}+(PC\cdot PA)^{3/2}\geq 12Rr^{2};}$^{[29]: p. 227}

${\displaystyle (PA\cdot PB)^{2}+(PB\cdot PC)^{2}+(PC\cdot PA)^{2}\geq 8(R+r)Rr^{2};}$^{[29]: p. 233}

${\displaystyle (PA\cdot PB)^{2}+(PB\cdot PC)^{2}+(PC\cdot PA)^{2}\geq 48r^{4};}$^{[29]: p. 233}

${\displaystyle (PA\cdot PB)^{2}+(PB\cdot PC)^{2}+(PC\cdot PA)^{2}\geq 6(7R-6r)r^{3}.}$^{[29]: p. 233}

Let ABC be a triangle, let G be its centroid, and let D, E, and F be the midpoints of BC, CA, and AB, respectively. For any point P in the plane of ABC:

${\displaystyle PA+PB+PC\leq 2(PD+PE+PF)+3PG.}$^[30]

Inradius, exradii, and circumradius

Inradius and circumradius

The Euler inequality for the circumradius R and the inradius r states that

${\displaystyle {\frac {R}{r}}\geq 2,}$

with equality only in the equilateral case.^{[31]: p. 198}

A stronger version^{[5]: p. 198} is

${\displaystyle {\frac {R}{r}}\geq {\frac {abc+a^{3}+b^{3}+c^{3}}{2abc}}\geq {\frac {a}{b}}+{\frac {b}{c}}+{\frac {c}{a}}-1\geq {\frac {2}{3}}\left({\frac {a}{b}}+{\frac {b}{c}}+{\frac {c}{a}}\right)\geq 2.}$

By comparison,^{[2]: p.183, #276.2}

${\displaystyle {\frac {r}{R}}\geq {\frac {4abc-a^{3}-b^{3}-c^{3}}{2abc}},}$

where the right side could be positive or negative.

Two other refinements of Euler's inequality are^{[2]: p.134, #3087}

${\displaystyle {\frac {R}{r}}\geq {\frac {(b+c)}{3a}}+{\frac {(c+a)}{3b}}+{\frac {(a+b)}{3c}}\geq 2}$

and

${\displaystyle \left({\frac {R}{r}}\right)^{3}\geq \left({\frac {a}{b}}+{\frac {b}{a}}\right)\left({\frac {b}{c}}+{\frac {c}{b}}\right)\left({\frac {c}{a}}+{\frac {a}{c}}\right)\geq 8.}$

Another symmetric inequality is^{[2]: p.125, #3004}

${\displaystyle {\frac {\left({\sqrt {a}}-{\sqrt {b}}\right)^{2}+\left({\sqrt {b}}-{\sqrt {c}}\right)^{2}+\left({\sqrt {c}}-{\sqrt {a}}\right)^{2}}{\left({\sqrt {a}}+{\sqrt {b}}+{\sqrt {c}}\right)^{2}}}\leq {\frac {4}{9}}\left({\frac {R}{r}}-2\right).}$

Moreover,

${\displaystyle {\frac {R}{r}}\geq {\frac {2(a^{2}+b^{2}+c^{2})}{ab+bc+ca}};}$^{[1]: 288}

${\displaystyle a^{3}+b^{3}+c^{3}\leq 8s(R^{2}-r^{2})}$

in terms of the semiperimeter s;^{[2]: p.20, #816}

${\displaystyle r(r+4R)\geq {\sqrt {3}}\cdot T}$

in terms of the area T;^{[5]: p. 201}

${\displaystyle s{\sqrt {3}}\leq r+4R}$ ^{[5]: p. 201}

and

${\displaystyle s^{2}\geq 16Rr-5r^{2}}$ ^{[2]: p.17#708}

in terms of the semiperimeter s; and

${\displaystyle {\begin{aligned}&2R^{2}+10Rr-r^{2}-2(R-2r){\sqrt {R^{2}-2Rr}}\leq s^{2}\\&\quad \leq 2R^{2}+10Rr-r^{2}+2(R-2r){\sqrt {R^{2}-2Rr}}\end{aligned}}}$

also in terms of the semiperimeter.^{[5]: p. 206 [7]: p. 99} Here the expression ${\displaystyle {\sqrt {R^{2}-2Rr}}=d}$ where d is the distance between the incenter and the circumcenter. In the latter double inequality, the first part holds with equality if and only if the triangle is isosceles with an apex angle of at least 60°, and the last part holds with equality if and only if the triangle is isosceles with an apex angle of at most 60°. Thus both are equalities if and only if the triangle is equilateral.^{[7]: Thm. 1}

We also have for any side a^[32]

${\displaystyle (R-d)^{2}-r^{2}\leq 4R^{2}r^{2}\left({\frac {(R+d)^{2}-r^{2}}{(R+d)^{4}}}\right)\leq {\frac {a^{2}}{4}}\leq Q\leq (R+d)^{2}-r^{2},}$

where ${\displaystyle Q=R^{2}}$ if the circumcenter is on or outside of the incircle and ${\displaystyle Q=4R^{2}r^{2}\left({\frac {(R-d)^{2}-r^{2}}{(R-d)^{4}}}\right)}$ if the circumcenter is inside the incircle. The circumcenter is inside the incircle if and only if^[32]

${\displaystyle {\frac {R}{r}}<{\sqrt {2}}+1.}$

Further,

${\displaystyle {\frac {9r}{2T}}\leq {\frac {1}{a}}+{\frac {1}{b}}+{\frac {1}{c}}\leq {\frac {9R}{4T}}.}$^{[1]: p. 291}

Blundon's inequality states that^{[5]: p. 206, [33][34]}

${\displaystyle s\leq (3{\sqrt {3}}-4)r+2R.}$

We also have, for all acute triangles,^[35]

${\displaystyle s>2R+r.}$

For incircle center I, let AI, BI, and CI extend beyond I to intersect the circumcircle at D, E, and F respectively. Then^{[2]: p.14, #644}

${\displaystyle {\frac {AI}{ID}}+{\frac {BI}{IE}}+{\frac {CI}{IF}}\geq 3.}$

In terms of the vertex angles we have ^{[2]: p.193, #342.6}

${\displaystyle \cos A\cdot \cos B\cdot \cos C\leq \left({\frac {r}{R{\sqrt {2}}}}\right)^{2}.}$

Denote as ${\displaystyle R_{A},R_{B},R_{C}}$ the tanradii of the triangle. Then^{[36]: Thm. 4}

${\displaystyle {\frac {4}{R}}\leq {\frac {1}{R_{A}}}+{\frac {1}{R_{B}}}+{\frac {1}{R_{C}}}\leq {\frac {2}{r}}}$

with equality only in the equilateral case, and ^[37]

${\displaystyle {\frac {9}{2}}r\leq R_{A}+R_{B}+R_{C}\leq 2R+{\frac {1}{2}}r}$

with equality only in the equilateral case.

Circumradius and other lengths

For the circumradius R we have^{[2]: p.101, #2625}

${\displaystyle 18R^{3}\geq (a^{2}+b^{2}+c^{2})R+abc{\sqrt {3}}}$

and^{[2] : p.35, #1130}

${\displaystyle a^{2/3}+b^{2/3}+c^{2/3}\leq 3^{7/4}R^{3/2}.}$

We also have^{[1]: pp. 287–90}

${\displaystyle a+b+c\leq 3{\sqrt {3}}\cdot R,}$

${\displaystyle 9R^{2}\geq a^{2}+b^{2}+c^{2},}$

${\displaystyle h_{a}+h_{b}+h_{c}\leq 3{\sqrt {3}}\cdot R}$

in terms of the altitudes,

${\displaystyle m_{a}^{2}+m_{b}^{2}+m_{c}^{2}\leq {\frac {27}{4}}R^{2}}$

in terms of the medians, and^{[2]: p.26, #957}

${\displaystyle {\frac {ab}{a+b}}+{\frac {bc}{b+c}}+{\frac {ca}{c+a}}\geq {\frac {2T}{R}}}$

in terms of the area.

Moreover, for circumcenter O, let lines AO, BO, and CO intersect the opposite sides BC, CA, and AB at U, V, and W respectively. Then^{[2]: p.17, #718}

${\displaystyle OU+OV+OW\geq {\frac {3}{2}}R.}$

For an acute triangle the distance between the circumcenter O and the orthocenter H satisfies^{[2]: p.26, #954}

${\displaystyle OH<R,}$

with the opposite inequality holding for an obtuse triangle.

The circumradius is at least twice the distance between the first and second Brocard points B₁ and B₂:^[38]

${\displaystyle R\geq 2B_{1}B_{2}.}$

Inradius, exradii, and other lengths

For the inradius r we have^{[1]: pp. 289–90}

${\displaystyle {\frac {1}{a}}+{\frac {1}{b}}+{\frac {1}{c}}\leq {\frac {\sqrt {3}}{2r}},}$

${\displaystyle 9r\leq h_{a}+h_{b}+h_{c}}$

in terms of the altitudes, and

${\displaystyle {\sqrt {r_{a}^{2}+r_{b}^{2}+r_{c}^{2}}}\geq 6r}$

in terms of the radii of the excircles. We additionally have

${\displaystyle {\sqrt {s}}({\sqrt {a}}+{\sqrt {b}}+{\sqrt {c}})\leq {\sqrt {2}}(r_{a}+r_{b}+r_{c})}$^{[2]: p.66, #1678}

and

${\displaystyle {\frac {abc}{r}}\geq {\frac {a^{3}}{r_{a}}}+{\frac {b^{3}}{r_{b}}}+{\frac {c^{3}}{r_{c}}}.}$^{[2]: p.183, #281.2}

The exradii and medians are related by^{[2]: p.66, #1680}

${\displaystyle {\frac {r_{a}r_{b}}{m_{a}m_{b}}}+{\frac {r_{b}r_{c}}{m_{b}m_{c}}}+{\frac {r_{c}r_{a}}{m_{c}m_{a}}}\geq 3.}$

In addition, for an acute triangle the distance between the incircle center I and orthocenter H satisfies^{[2]: p.26, #954}

${\displaystyle IH<r{\sqrt {2}},}$

with the reverse inequality for an obtuse triangle.

Also, an acute triangle satisfies^{[2]: p.26, #954}

${\displaystyle r^{2}+r_{a}^{2}+r_{b}^{2}+r_{c}^{2}<8R^{2},}$

in terms of the circumradius R, again with the reverse inequality holding for an obtuse triangle.

If the internal angle bisectors of angles A, B, C meet the opposite sides at U, V, W then^{[2]: p.215, 32nd IMO, #1}

${\displaystyle {\frac {1}{4}}<{\frac {AI\cdot BI\cdot CI}{AU\cdot BV\cdot CW}}\leq {\frac {8}{27}}.}$

If the internal angle bisectors through incenter I extend to meet the circumcircle at X, Y and Z then ^{[2]: p.181, #264.4}

${\displaystyle {\frac {1}{IX}}+{\frac {1}{IY}}+{\frac {1}{IZ}}\geq {\frac {3}{R}}}$

for circumradius R, and^{[2]: p.181, #264.4 [2]: p.45, #1282}

${\displaystyle 0\leq (IX-IA)+(IY-IB)+(IZ-IC)\leq 2(R-2r).}$

If the incircle is tangent to the sides at D, E, F, then^{[2]: p.115, #2875}

${\displaystyle EF^{2}+FD^{2}+DE^{2}\leq {\frac {s^{2}}{3}}}$

for semiperimeter s.

Inscribed figures

Inscribed hexagon

If a tangential hexagon is formed by drawing three segments tangent to a triangle's incircle and parallel to a side, so that the hexagon is inscribed in the triangle with its other three sides coinciding with parts of the triangle's sides, then^{[2]: p.42, #1245}

${\displaystyle {\text{Perimeter of hexagon}}\leq {\frac {2}{3}}({\text{Perimeter of triangle}}).}$

Inscribed triangle

If three points D, E, F on the respective sides AB, BC, and CA of a reference triangle ABC are the vertices of an inscribed triangle, which thereby partitions the reference triangle into four triangles, then the area of the inscribed triangle is greater than the area of at least one of the other interior triangles, unless the vertices of the inscribed triangle are at the midpoints of the sides of the reference triangle (in which case the inscribed triangle is the medial triangle and all four interior triangles have equal areas):^{[9]: p.137}

${\displaystyle {\text{Area(DEF)}}\geq \min({\text{Area(BED), Area(CFE), Area(ADF)}}).}$

Inscribed squares

An acute triangle has three inscribed squares, each with one side coinciding with part of a side of the triangle and with the square's other two vertices on the remaining two sides of the triangle. (A right triangle has only two distinct inscribed squares.) If one of these squares has side length x_a and another has side length x_b with x_a < x_b, then^{[39]: p. 115}

${\displaystyle 1\geq {\frac {x_{a}}{x_{b}}}\geq {\frac {2{\sqrt {2}}}{3}}\approx 0.94.}$

Moreover, for any square inscribed in any triangle we have^{[2]: p.18, #729 [39]}

${\displaystyle {\frac {\text{Area of triangle}}{\text{Area of inscribed square}}}\geq 2.}$

Euler line

A triangle's Euler line goes through its orthocenter, its circumcenter, and its centroid, but does not go through its incenter unless the triangle is isosceles.^{[16]: p.231} For all non-isosceles triangles, the distance d from the incenter to the Euler line satisfies the following inequalities in terms of the triangle's longest median v, its longest side u, and its semiperimeter s:^{[16]: p. 234, Propos.5}

${\displaystyle {\frac {d}{s}}<{\frac {d}{u}}<{\frac {d}{v}}<{\frac {1}{3}}.}$

For all of these ratios, the upper bound of 1/3 is the tightest possible.^{[16]: p.235, Thm.6}

Right triangle

In right triangles the legs a and b and the hypotenuse c obey the following, with equality only in the isosceles case:^{[1]: p. 280}

${\displaystyle a+b\leq c{\sqrt {2}}.}$

In terms of the inradius, the hypotenuse obeys^{[1]: p. 281}

${\displaystyle 2r\leq c({\sqrt {2}}-1),}$

and in terms of the altitude from the hypotenuse the legs obey^{[1]: p. 282}

${\displaystyle h_{c}\leq {\frac {\sqrt {2}}{4}}(a+b).}$

Isosceles triangle

If the two equal sides of an isosceles triangle have length a and the other side has length c, then the internal angle bisector t from one of the two equal-angled vertices satisfies^{[2]: p.169, #${\displaystyle \eta }$44}

${\displaystyle {\frac {2ac}{a+c}}>t>{\frac {ac{\sqrt {2}}}{a+c}}.}$

Equilateral triangle

For any point P in the plane of an equilateral triangle ABC, the distances of P from the vertices, PA, PB, and PC, are such that, unless P is on the triangle's circumcircle, they obey the basic triangle inequality and thus can themselves form the sides of a triangle:^{[1]: p. 279}

However, when P is on the circumcircle the sum of the distances from P to the nearest two vertices exactly equals the distance to the farthest vertex.

A triangle is equilateral if and only if, for every point P in the plane, with distances PD, PE, and PF to the triangle's sides and distances PA, PB, and PC to its vertices,^{[2]: p.178, #235.4}

Two triangles

Pedoe's inequality for two triangles, one with sides a, b, and c and area T, and the other with sides d, e, and f and area S, states that

${\displaystyle d^{2}(b^{2}+c^{2}-a^{2})+e^{2}(a^{2}+c^{2}-b^{2})+f^{2}(a^{2}+b^{2}-c^{2})\geq 16TS,}$

with equality if and only if the two triangles are similar.

The hinge theorem or open-mouth theorem states that if two sides of one triangle are congruent to two sides of another triangle, and the included angle of the first is larger than the included angle of the second, then the third side of the first triangle is longer than the third side of the second triangle. That is, in triangles ABC and DEF with sides a, b, c, and d, e, f respectively (with a opposite A etc.), if a = d and b = e and angle C > angle F, then

${\displaystyle c>f.}$

The converse also holds: if c > f, then C > F.

The angles in any two triangles ABC and DEF are related in terms of the cotangent function according to^[6]

${\displaystyle \cot A(\cot E+\cot F)+\cot B(\cot F+\cot D)+\cot C(\cot D+\cot E)\geq 2.}$

Non-Euclidean triangles

In a triangle on the surface of a sphere, as well as in elliptic geometry,

${\displaystyle \angle A+\angle B+\angle C>180^{\circ }.}$

This inequality is reversed for hyperbolic triangles.

See also

• List of inequalities
• List of triangle topics
• Quadrilateral § Inequalities
• Quadrilateral § Maximum and minimum properties

References