Inequalities in Triangle
Introduction
The elements of a triangle - sides, angles, inradius, circumradius, altitudes, area, etc. - satisfy a great number of relations, like the Law of Sines or Heron's formula. Triangle elements are also bound by inequalities, foremost of which is the Triangle Inequality (inequalities, actually.) Below I collect several inequalities that bind triangle elements. (As in the sister page on the identities between such quantities, I'll use capital letters, $A,B,C$ to denote both vertices and the corresponding inner angles of a triangle.)
$\sin\frac{A}{2}\le\frac{a}{2\sqrt{bc}}$
By the Law of Cosines,
$a^{2}=b^{2}+c^{2}-2bc\cos(A)=(b-c)^{2}+4bc\,\sin^{2}(\frac{A}{2}).$
It follows that $a^{2}\ge 4bc\,\sin^{2}(\frac{A}{2}),$ $a\ge 2\sqrt{bc}\sin\frac{A}{2}.$
$\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}\le\frac{1}{8}$
Apply the previous inequality cyclically:
$\begin{align} \sin\frac{A}{2} &\le\frac{a}{2\sqrt{bc}},\\ \sin\frac{B}{2} &\le\frac{b}{2\sqrt{ca}},\\ \sin\frac{C}{2} &\le\frac{c}{2\sqrt{ab}}. \end{align}$
Multiply the three:
$\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}\le\frac{abc}{8\sqrt{a^{2}b^{2}c^{2}}}=\frac{1}{8}.$
$\cos (A)+\cos (B)+\cos (C)\le\frac{3}{2}$
By the addition formulas,
$\begin{align} \cos(A)+\cos(B) &= 2\cos\frac{A+B}{2}\cos\frac{A-B}{2}\\ &=2\cos\frac{\pi -C}{2}\cos\frac{A-B}{2}\\ &=2\sin\frac{C}{2}\cos\frac{A-B}{2}. \end{align}$
Also, $\cos(C)=1-2\,\sin^{2}\frac{C}{2},$ so that
$\begin{align} \cos (A)+\cos (B)+\cos (C) &=1+2\,\sin\frac{C}{2}\bigg(\cos\frac{A-B}{2}-\sin\frac{C}{2}\bigg)\\ &\le 1+2\,\sin\frac{C}{2}\bigg(1-\sin\frac{C}{2}\bigg)\\ &\le 1+2\cdot\frac{1}{4}\\ &\le\frac{3}{2}, \end{align}$
because for any real $x,$ $x(1-x)\le\frac{1}{4},$ with equality only when $x=\frac{1}{2}.$ The trigonometric inequality becomes equality only when $A=B=C=60^{\circ}.$
$\cos A\cos B\cos C\le\frac{1}{8}$
With the AM-GM inequality, that's a direct consequences of the previous inequality (for acute triangles):
$\displaystyle \frac{3}{2}\ge\sum_{cycl}\cos A\ge 3\sqrt[3]{\cos A\cos B\cos C}$
so that $\displaystyle \left(\frac{1}{2}\right)^3\ge\cos A\cos B\cos C.$
For obtuse triangles the inequality is trivial.
$\displaystyle\sin\frac{A}{2}+\sin\frac{B}{2}+\sin\frac{C}{2}\le\frac{3}{2}.$
Observe that, if $\displaystyle\alpha=\frac{\pi-A}{2},\;$ $\displaystyle\beta=\frac{\pi-B}{2},\;$ $\displaystyle\gamma=\frac{\pi-C}{2},\;$ then $\alpha+\beta+\gamma=\pi,\;$ so that there is a triangle with angles $\alpha,\;$ $\beta,\;$ and $\gamma.\;$ Thus
$\displaystyle\begin{align} \sin\frac{A}{2}+\sin\frac{B}{2}+\cos\frac{C}{2} &= \cos\frac{\pi-A}{2}+\cos\frac{\pi-B}{2}+\cos\frac{\pi-C}{2}\\ &=\cos\alpha+\cos\beta+\cos\gamma\\ &\le\frac{3}{2} \end{align}$
$\cos(\frac{A}{2})\cos(\frac{B}{2})\cos(\frac{C}{2})\le\frac{3\sqrt{3}}{8}$
We already know that $\cos (A)+\cos (B)+\cos (C)\le\frac{3}{2}.\;$ Using $\displaystyle\cos\alpha =2\cos^2\frac{\alpha}{2}-1,\;$ $\displaystyle\cos^2\frac{A}{2}+\cos^2\frac{B}{2}+\cos^2\frac{C}{2}\le\frac{9}{4}.\;$ From the AM-GM inequality then
$\displaystyle\begin{align} \cos\frac{A}{2}\cos\frac{B}{2}\cos\frac{C}{2}&\le\left[\frac{1}{3}\left(\cos^2\frac{A}{2}+\cos^2\frac{B}{2}+\cos^2\frac{C}{2}\right)\right]^{\frac{3}{2}}\\ &\le\left[\frac{3}{4}\right]^{\frac{3}{2}}\\ &=\frac{3\sqrt{3}}{8}. \end{align}$
$m^{2}_{a}+m^{2}_{b}+m^{2}_{c}\le\frac{27}{4}R^{2}$
By Lagrange's theorem of moments, if $G$ is the centroid of $\Delta ABC$ and $O$ is the circumcenter,
$\begin{align} 3R^{2} &= OA^{2}+OB^{2}+OC^{2}\\ &= GA^{2}+GB^{2}+GC^{2}+3GO^{2}\\ &\ge GA^{2}+GB^{2}+GC^{2}\\ &=\frac{4}{9}(m^{2}_{a}+m^{2}_{b}+m^{2}_{c}). \end{align}$
$a^{2}+b^{2}+c^{2}\le 9R^{2}$
It is easy to check that $4(m_a^2+m_b^2+m_c^2)=3(a^2+b^2+c^2).\,$ Then from the previous inequality, we directly get $a^{2}+b^{2}+c^{2}\le 9R^{2}.$
$6\sqrt{3}r\le a+b+c$
From the isoperimetric theorem for triangles,
$\displaystyle \left(\frac{a+b+c}{2}\right)^2 \ge 3\sqrt{3}S=3\sqrt{3}\frac{a+b+c}{2}r,$
so that $\displaystyle \frac{a+b+c}{2}\ge 3\sqrt{3}r,\,$ which is the required inequality.
This inequality and the next one are usually attributed to D. S. Mitrinović.
$a+b+c\le 3\sqrt{3}R$
$\begin{align} 4m^{2}_{a}=2(b^{2}+c^{2})-a^{2},\\ 4m^{2}_{b}=2(c^{2}+a^{2})-b^{2},\\ 4m^{2}_{c}=2(a^{2}+b^{2})-c^{2},\\ \end{align}$
such that $4(m^{2}_{a}+m^{2}_{b}+m^{2}_{c})=3(a^{2}+b^{2}+c^{2}).\;$ Note that
$\begin{align} 2(ab+bc+ca)&=2(a^{2}+b^{2}+c^{2})-((a-b)^{2}+(b-c)^{2}+(c-a)^{2})\\ &\le 2(a^{2}+b^{2}+c^{2}) \end{align}$
implying,
$\begin{align} 4(m^{2}_{a}+m^{2}_{b}+m^{2}_{c}) &= 3(a^{2}+b^{2}+c^{2})\\ &\ge (a^{2}+b^{2}+c^{2}) + 2(ab+bc+ca)\\ &= (a + b+ c)^{2}. \end{align}$
Now, taking into account already proved $m^{2}_{a}+m^{2}_{b}+m^{2}_{c}\le\frac{27}{4}R^{2},\,$ we get $a+b+c\le 3\sqrt{3}R,\,$ as required.
This inequality and the previous one are usually attributed to D. S. Mitrinović.
$\sin (A)+\sin (B)+\sin (C)\le\frac{3\sqrt{3}}{2}$
Since $a=2R\,\sin (A),$ etc., this is equivalent to the previous inequality.
$\displaystyle\cos\frac{A}{2}+\cos\frac{B}{2}+\cos\frac{C}{2}\le\frac{3\sqrt{3}}{2}$
Observe that, if $\displaystyle\alpha=\frac{\pi-A}{2},\;$ $\displaystyle\beta=\frac{\pi-B}{2},\;$ $\displaystyle\gamma=\frac{\pi-C}{2},\;$ then $\alpha+\beta+\gamma=\pi,\;$ so that there is a triangle with angles $\alpha,\;$ $\beta,\;$ and $\gamma.\;$ Thus
$\displaystyle\begin{align} \cos\frac{A}{2}+\cos\frac{B}{2}+\cos\frac{C}{2} &= \sin\frac{\pi-A}{2}+\sin\frac{\pi-B}{2}+\sin\frac{\pi-C}{2}\\ &=\sin\alpha+\sin\beta+\sin\gamma\\ &\le\frac{3\sqrt{3}}{2} \end{align}$
$\sin (A)\,\sin (B)\,\sin (C)\le\frac{3\sqrt{3}}{8}$
We'll use the Arithmetic Mean - Geometric Mean inequality: for $x,y,z\gt 0,$ $\displaystyle\frac{x+y+z}{3}\ge \sqrt[3]{xyz}.$ The inequality shows that $x+y+z\le\frac{3\sqrt{3}}{2}$ implies $\sqrt[3]{xyz}\le\frac{\sqrt{3}}{2},$ or $xyz\le\frac{3\sqrt{3}}{8}.$ Since, as we already seen $\sin (A)+\sin (B)+\sin (C)\le\frac{3\sqrt{3}}{2},\;$ the required inequality follows.
$a^3b+b^3c+c^3a-a^2b^2-b^2c^2-c^2a^2\ge 0$
I have placed the the above in a separate file.
$R\ge 2r$
This is known as (another) Euler's inequality. The inequality is an immediate consequence of Euler's identity, $OI^2 = R^2 - 2Rr.$ There are additional proofs, so I start a separate page in case I come across more of them.
$m_a l_a+m_b l_b+m_c l_c\ge p^{2}$
I have placed the the above in a separate file.
$\displaystyle\frac{a(b+c)}{\displaystyle bc\cos^2\frac{A}{2}}+\frac{b(c+a)}{\displaystyle ca\cos^2\frac{B}{2}}+\frac{c(a+b)}{\displaystyle ab\cos^2\frac{C}{2}}\ge 8$
I have placed the proof in a separate file.
$\displaystyle\tan\frac{A}{2}+\tan\frac{B}{2}+\tan\frac{C}{2}\ge\sqrt{3}$
$\displaystyle\tan\frac{A}{2}\tan\frac{B}{2}+\tan\frac{B}{2}\tan\frac{C}{2}+\tan\frac{C}{2}\tan\frac{A}{2}=1.$
By rearrangemnt,
$\displaystyle\tan^2\frac{A}{2}+\tan^2\frac{B}{2}+\tan^2\frac{C}{2}\ge 1.$
Multiplying the first by $2\;$ and adding up gives
$\displaystyle\left(\tan\frac{A}{2}+\tan\frac{B}{2}+\tan\frac{C}{2}\right)^2\ge 3.$
$\displaystyle m_a+m_b+m_c\le 4R+r$
This is known as Leuenberger's inequality - one of a few that look deceptively simple. I have placed the proof in a separate file.
Acknowledgment
The books Problems in Planimetry (especially the second volume) by V. V. Prasolov (Nauka, Moscow, 1986 (Russian)) are an excellent starting point as is the book by O. Bottema et al Geometric Inequalities that is available on the internet as a pdf file.
- Inequalities in Triangle
- Padoa's Inequality
$(abc\ge (a+b-c)(b+c-a)(c+a-b))$
- Refinement of Padoa's Inequality $\left(\displaystyle \prod_{cycl}(a+b-c)\le 2\min_{cycl}\{a\cdot\frac{b^2c^2}{b^2+c^2}\}\le 2\max_{cycl}\{a\cdot\frac{b^2c^2}{b^2+c^2}\}\le abc\right)$
- Erdos-Mordell Inequality $(OA+OB+OC\ge 2(OP+OQ+OR) )$
- From Triangle Inequality to Inequality in Triangle $(\max\{A,B,C,D,E,F\}\le a^ab^bc^c)$
- Area Inequalities in Triangle $([\Delta NAP]\le\frac{1}{4}[\Delta ABC])$
- Area Inequality in Three Triangles $\displaystyle 2(\sqrt{S_1}+\sqrt{S_2}+\sqrt{S_3})^2 \lt \sum_{cycl}a_1^2 + \sum_{cycl}a_2^2 + \sum_{cycl}a_3^2.$
- Area Inequality in Triangle II $([PBF]\le\frac{1}{6})$
- An Inequality in Triangle $(a^3b+b^3c+c^3a-a^2b^2-b^2c^2-c^2a^2\ge 0)$
- An Inequality in Triangle, II $(m_al_a+m_bl_b+m_cl_c\ge p^2)$
- An Inequality in Triangle III $\displaystyle\left(\frac{a(b+c)}{bc\cdot\cos^2\frac{A}{2}}+\frac{b(c+a)}{ca\cdot\cos^2\frac{B}{2}}+\frac{c(a+b)}{ab\cdot\cos^2\frac{C}{2}}\ge 8\right)$
- An Inequality in Triangle IV $\left(\begin{align}\sqrt{2}&\left[\sqrt{p(p-a)}+\sqrt{p(p-b)}+\sqrt{p(p-c)}\right]\\ &\le\sqrt{p^2-m_a^2}+\sqrt{p^2-m_b^2}+\sqrt{p^2-m_c^2}, \end{align}\right)$
- An Inequality in Triangle, V $(m_am_bm_c\ge r_ar_br_c)$
- An Inequality in Triangle, VI $\displaystyle\left(\frac{h_a\cdot h_b}{h_a+h_b}\right)\lt h_c\left(\frac{h_a\cdot h_b}{|h_a-h_b|}\right)$
- An Inequality in Triangle, VII $\displaystyle\left(\left(\sum_{cycl}\frac{m_a^2}{m_b^2}\right)\left(\sum_{cycl}x^2\right)+2\left(\sum_{cycl}\frac{m_a}{m_c}\right)\left(\sum_{cycl}xy\right)\ge 0\right)$
- An Inequality in Triangle, VIII $\displaystyle\left(\sum_{cycl}\frac{5a^2-b^2-c^2}{\sqrt{m_bm_c}}\le 4\sum_{cycl}m_a\right)$
- An Inequality in Triangle, IX $\displaystyle\left( 27\prod_{cycl}IA'\cdot HA''\le\frac{1}{27}\prod_{cycl}\ell_ah_a\right)$
- An Inequality in Triangle, X $\displaystyle\left(\frac{1}{r^2}\sum_{cycl}a^3\cos B\cos C\ge 16\left(\sum_{cycl}\sin A\right)\left(\sum_{cycl}\cos^2 A\right)\right)$
- An Inequality in Triangle XI $\left(3(a^2+b^2+c^2)\lt 4(am_c+bm_a+cm_b)\right)$
- Inequality In Triangle: Sides and Angle Bisectors $\left(\displaystyle a+b+c \ge \frac{2\sqrt{3}}{3}(l_a+l_b+l_c)\right)$
- Weitzenböck's inequality
$(a^2 + b^2 + c^2 \ge 4\sqrt{3}S)$
- Two Refinements of the Ionescu-Weitzenbock Inequality $(a^2+b^2+c^2\ge 2\sqrt{3}\max\{am_a,bm_b,cm_c\})$
- Another Refinement of the Ionescu-Weitzenbock Inequality $(\displaystyle a^2+b^2+c^2-4\sqrt{3}S\ge 2\sqrt{3}(m_a^2-h_a^2))$
- Early Refinement of the Ionescu-Weitzenbock Inequality $(\displaystyle a^2+b^2+c^2-(a-b)^2-(b-c)^2-(c-a)^2\ge 4\sqrt{3}S)$
- Weitzenbock by Sanchez $([ABC](1+\sqrt{3}]\le [ANBMCP])$
- An Inequality In Triangle That Involves the Four Basic Centers $\left(\displaystyle \sum_{cycl}(AH+2\cdot AI+3\cdot AO+4\cdot AG)\ge 60r\right)$
- An Inequality in Acute Triangle, Courtesy of Ceva's Theorem $\displaystyle\left(AB'\cdot BC'\cdot CA'+AB''\cdot BC''\cdot CA''+AB'''\cdot BC'''\cdot CA'''\le \frac{3}{8}abc\right)$
- Problem 4020 from Crux Mathematicorum $([MNP] \le [DEF])$
- A Two-Triangle Inequality $(a^2(-a'^2+b'^2+c'^2)+b^2(a'^2-b'^2+c'^2)+c^2(a'^2+b'^2-c'^2)\ge 16KK')$
- A Two-Triangle Inequality II $\left(\displaystyle\frac{a+b+c}{3\sqrt{3}R}\le\frac{\displaystyle\cos\frac{A}{2}+\cos\frac{B}{2}+\cos\frac{C}{2}}{\displaystyle\cos\frac{A'}{2}+\cos\frac{B'}{2}+\cos\frac{C'}{2}}\le\frac{3\sqrt{3}R'}{a'+b'+c'}\right)$
- Points on Incircle: Another Look $\left(\displaystyle 5r\le\frac{PA^2}{h_a}+\frac{PB^2}{h_b}+\frac{PC^2}{h_c}\le\frac{5}{2}R\right)$
- An All-Inclusive Inequality $\left(\displaystyle\frac{m_am_bm_c}{r_ar_br_c}+\frac{\ell_a\ell_b\ell_c}{h_ah_bh_c}\leq \frac{R}{r}\right)$
- An All-Inclusive Inequality II $\left(\displaystyle\left(\sum_{cycl}\sqrt{\frac{m_a}{\ell_a}}\right)\left(\sum_{cycl}\sqrt{\frac{m_a}{h_a}}\right)\le\frac{9R}{2r}\right)$
- Inequality with Roots, Squares and the Area $(\displaystyle \sqrt{2}(PA+PB+PC)\ge\sqrt{a^2+b^2+c^2+4\sqrt{3}S})$
- A One-Sided Inequality in Triangle $\left(\displaystyle BA'\cdot CB'\cdot AC' + BA''\cdot CB''\cdot AC'' + BA'''\cdot CB'''\cdot AC''' \lt \frac{3abc}{8}\right)$
- Dan Sitaru's Inequality with Tangents $(\displaystyle\sum_{cycl}\sqrt[3]{\tan A}\sqrt[3]{\tan B}(\sqrt[3]{\tan A}+\sqrt[3]{\tan B})\le 2\tan A\tan B\tan C)$
- Dan Sitaru's Inequality with Tangents II $(\displaystyle \sum_{cycl}\tan A\tan B + 45^{\circ}\le 2\tan^2A\tan^2B\tan^2C)$
- Dan Sitaru's Inequality with Roots and Powers $(\displaystyle (\sqrt{a}+\sqrt{b}+\sqrt{c})^4(\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c})^6\ge 2^43^9S^2)$
- A Cyclic Inequality in Triangle $\left(\displaystyle\sum_{cycl}\frac{a^3(2s-a)}{b(2s-b)}\ge\frac{27a^2b^2c^2}{s^2}\right)$
- A Cyclic Inequality in Triangle II $\left(\displaystyle\sqrt{abc}\left(\frac{a^2}{\sqrt{b}}+\frac{b^2}{\sqrt{c}}+\frac{c^2}{\sqrt{a}}\right)^2\ge 16(\sqrt{a}+\sqrt{b}+\sqrt{c})S^2\right)$
- Inequality with Cubes and Cube Roots $(\displaystyle\sum_{cycl}(\sqrt[3]{a}+\sqrt[3]{b}-\sqrt[3]{c})^3\ge\sqrt[3]{3a}+\sqrt[3]{3b}+\sqrt[3]{3c}-2)$
- Tangent, Cotangent and Square Roots Inequality $(\displaystyle\left(\sum_{cycl}\sqrt{\cot A\cot B}\right)\left(\sum_{cycl}\sqrt{\tan A\cot B}\right)\ge 3\sqrt{3})$
- An Inequality with Sines $\left(\displaystyle \prod_{cycl}\left(\frac{2}{\sin A}-1\right)\ge\left(\frac{6}{\sin A+\sin B+\sin C}-1\right)^3\right)$
- An Inequality in Triangle, with Sines II $\left(\displaystyle \left(\sum_{cycl}\frac{\sin A}{\sin B}\right)\left(\sum_{cycl}\frac{\sin A}{\sin^2 B}\right)\left(\sum_{cycl}\frac{\sin A}{\sin^3 B}\right)\ge 24\sqrt{3}\right)$ $\left(\displaystyle\sin^22A+\sin^22B+\sin^22C\le\sin^2A+\sin^2B+\sin^2C\right)$
- An Inequality with Tangents and Cotangents $\left(\displaystyle\prod_{cycl}\left(\tan\frac{A}{2}\tan\frac{B}{2}+\cot\frac{A}{2}\cot\frac{B}{2}\right)\ge\frac{1000}{27}\right)$
- An Inequality with Sides and Medians $(2am_a\le bm_c+cm_b)$
- An Inequality in Triangle with Sides and Medians II $\left(\displaystyle 16\sum \Bigr(\frac{m_a}{m_c}+\frac{m_b}{m_c}\Bigr)^4\gt 81\Biggl(\Bigr(\frac{a}{m_a}\Bigr)^4+\Bigr(\frac{b}{m_b}\Bigr)^4+\Bigr(\frac{c}{m_c}\Bigr)^4\Biggl)\right)$
- An Inequality with Sin, Cos, Tan, Cot, and Some $(2S^2\displaystyle\sum_{cycl}(\sin A+\cos A+\tan A+\cot A)\gt 81\pi R^4\prod_{cycl}\cos A)$
- Leo Giugiuc's Second Lemma And Applications $(3(a+b)\gt 2(m_a+m_b))$
- An Inequality with Arctangents in Triangle $\left(\displaystyle\frac{a^3\cos^3A}{\arctan\frac{1}{2}}+\frac{b^3\cos^3B}{\arctan\frac{1}{5}}+\frac{c^3\cos^3C}{\arctan\frac{1}{8}}\ge\frac{32r^3s^3}{3\pi R^3}\right)$
- An Inequality in Triangle with Roots and Circumradius $(\displaystyle a\sqrt{b}+b\sqrt{c}+c\sqrt{a}\le 3R\sqrt{2s})$
- An Inequality for the Cevians through Circumcenter $\left(\displaystyle \frac{A_1O}{OA}+\frac{B_1O}{OB}+\frac{C_1O}{OC}=\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y}\ge\frac{3}{2}\right)$
- An Inequality with Powers of Six $(\displaystyle a^6+b^6+c^6\ge 8r^2s\sum_{cycl}\frac{a^5}{b^2-bc+c^2})$
- Adil Abdullayev's Inequality With Roots and Powers $\left(\displaystyle a^2+b^2+c^2\ge 4S\cdot\sqrt[4]{(a^2+b^2+c^2)\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)}\right)$
- Marian Cucoanes' Inequality With Roots and Powers $\left(\displaystyle \small{a^2+b^2+c^2\ge 4S\cdot\sqrt{\frac{1}{2}(a^2+b^2+c^2)\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)-\frac{3}{2}}}\right)$
- Marian Dinca's Inequality $\left(\displaystyle m_a\le\frac{s}{\sqrt{3}}\right)$
- An Inequality for the Cevians through Spieker Point via Brocard Angle $(a^2b^2+b^2c^2+c^2a^2 ≥ 2s(AC'\cdot BA'\cdot CB' + AB'\cdot BC'\cdot CA'))$
- Hung Nguyen Viet's Inequality with Radicals and Chebyshev $\left(\displaystyle\sum_{cycl}(a-\sqrt{bc})\sin\frac{A}{2}\ge 0\right)$
- An Inequality in Triangle, Mostly with the Medians $(\displaystyle \prod_{cycl}(5m_a+3m_b)(3m_a+5m_b)\lt 64\prod_{cycl}(2s+a)^2)$
- An Inequality in Triangle with Altitudes, Medians And Symmedians $\left(\displaystyle \frac{A_2A_3}{A_1A_2}\cdot\frac{B_2B_3}{B_1B_2}\cdot\frac{C_2C_3}{C_1C_2}\gt\prod_{cycl}\frac{(a+b-c)^2}{2a^2+2b^2-c^2}\right)$
- An Inequality with Altitudes and Medians $(\sqrt{3}\max (h_a,h_b,h_c)\ge s \ge \sqrt{3}\min (m_a,m_b,m_c))$
- An Inequality with Altitudes and Angle Bisectors $(\max (h_a,h_b,h_c)\ge \min (\ell_a,\ell_b,\ell_c))$
- Leo Giugiuc's Inequality for the Medians $(m_a+m_b+m_c\le\sqrt{3s^2+\frac{3}{4}[(a-b)^2+(b-c)^2+(c-a)^2]})$
- An Inequality in Triangle with Medians, Sides and Circumradius $(\displaystyle m_a\ge\frac{b^2+c^2}{4R})$
- An Inequality In Triangle with Sines of Half-Angles $\displaystyle\left( \sin\frac{A}{2}+\sin\frac{B}{2}+\sin\frac{C}{2}\le 2\sum_{cycl}\frac{a}{(\sqrt[3]{b}+\sqrt[3]{c})(\sqrt[3]{b^2}+\sqrt[3]{c^2})}\right)$
- An Inequality in Triangle with the Sines of Half-Angles and Cube Roots $\displaystyle\left( 2\sum_{cycl} \left(\frac{a}{b}+\frac{b}{a}\right)\sin^2 \frac{C}{2}\geq \sqrt[3]{abc}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\right)$
- An Inequality in Triangle with the Circumradius, Inradius and Angle Bisectors $(R+r\ge\min (\ell_a,\ell_b,\ell_c))$
- An Inequality in Triangle with Differences of the Medians $\displaystyle\left(\frac{8(m_a-m_b)(m_b-m_c)(m_c-m_a)}{(b-a)(c-b)(a-c)}>\frac{27abc}{(a+2s)(b+2s)(c+2s)}\right)$
- Problem 4087 from Crux Mathematicorum $( m_a(b+c)+2m_a^2\ge 4S\sin A)$
- An Inequality with Inradius and Circumradii $\left(\displaystyle (R_a+R_b+R_c)\left(\frac{R_a}{R_bR_c}+\frac{R_b}{R_cR_a}+\frac{R_c}{R_aR_b}\right)\ge 12-\frac{6r}{R}\right)$
- An Inequality in Triangle, with Integrals $\left(\displaystyle 4\sum_{cycl}\sin^2\frac{a}{2}+\pi\sum_{cycl}\int_{0}^{a}\cos (\sin x)dx\ge \pi^2\right)$
- An Inequality in Triangle, with Sines $\displaystyle\left(\prod_{cycl}\left(\frac{2}{\sin A}-1\right)\ge\left(\frac{6}{\sin A+\sin B+\sin C}-1\right)^3\right)$
- An Inequality in Triangle, with Sides and Sums $\left(\displaystyle \frac{a(2s-a)}{4(s-a)}+\frac{a(2s-b)}{4(s-b)}+\frac{a(2s-c)}{4(s-c)}\ge a+b+c\right)$
- An Inequality with a Variety of Circumradii $\left(\displaystyle \frac{R_a^2}{R_b}+\frac{R_b^2}{R_c}+\frac{R_c^2}{R_a}\geq 3R\right)$
- Dorin Marghidanu's Inequality with Maximum Side $\left(\displaystyle h_a \le \frac{p}{\sqrt{3}}\right)$
- An Inequality with Circumradii And Distances to the Vertices $\displaystyle\left(\frac{MB\cdot MC}{R_a}+\frac{MC\cdot MA}{R_b}+\frac{MA\cdot MB}{R_c}\le MA+MB+MC\right)$
- An inequality with Cosines and a Sine $\displaystyle\left(\cos A+4\cos B+4\sin\frac{C}{2}\le 9\cos\frac{\pi+B-C}{3}\right)$
- An inequality with Three Points $\displaystyle\left(\sum_{P\in\{O,I,G\}}\sum_{cycl}\left(\frac{[\Delta APB]}{[\Delta ABC]}+\frac{[\Delta ABC]}{[\Delta APB]}\right)^2\ge 100\right)$
- An Inequality with One Tangent and Six Sines $\left(\displaystyle \frac{\tan A}{\sin B+5\sin C}+\frac{\tan B}{\sin C+5\sin A}+\frac{\tan C}{\sin A+5\sin B}\gt\frac{1}{2}\right)$
- An Inequality with Tangents and Sides $\left(\displaystyle \frac{a^2}{\tan B+\tan C}+\frac{b^2}{\tan C+\tan A}+\frac{c^2}{\tan A+\tan B}\leq sR\right)$
- An Inequality with Sides, Altitudes, Angle Bisectors and Medians $\left(\displaystyle \left(\frac{h_b}{m_a}+\frac{h_c}{m_b}+\frac{h_a}{m_c}\right)\left(\frac{h_b}{\ell_a}+\frac{h_c}{\ell_b}+\frac{h_a}{\ell_c}\right)\ge\left(\frac{b}{a}+\frac{c}{b}+\frac{a}{c}\right)^2\right)$
- An Inequality of the Areas of Triangles Formed by Circumcenter And Orthocenter $\left(\displaystyle \sum_{cycl}\sqrt[n]{[\Delta OAB]}\ge\sum_{cycl}\sqrt[n]{[\Delta HAB]}\right)$
- An Inequality with Angle Trisectors $\left(\displaystyle \frac{AE}{AB}+\frac{AF}{AC}\lt 2\right)$
- An Inequality for Sides and Area $\left(\displaystyle \sum_{cycl}\frac{(a^2-ab+b^2)^2}{a^2+4ab+b^2}\ge\frac{2S}{\sqrt{3}}\right)$
- A Cyclic Inequality in Triangle for Integer Powers $\left(\displaystyle \sum_{cycl}\frac{a^{n+1}}{b+c-a}\ge\sum_{cycl}a^n\right)$
- R and r When G on Incircle $(8R\ge 25r)$
- An Inequality for Cevians And The Ratio of Circumradius and Inradius $\left(\displaystyle \frac{XB}{XY}\cdot\frac{YC}{YZ}\cdot\frac{ZA}{ZX}\le\frac{R}{2r}\right)$
- Centroid on the Incircle in Right Triangle $\left(648Rr\ge 25 (a^2+b^2+c^2)\right)$
- An Inequality with Cotangents And the Circumradius $\left(\displaystyle \sum_{cycl}a^2\cot B\cot C\le 4R^2\right)$
- An Inequality with Inradius and Excenters $\left(\displaystyle \sum_{cycl}\frac{1}{II_a^2}+\sum_{cycl}\frac{1}{I_aI_b^2}\le\frac{1}{4r^2}\right)$
- An Inequality with Inradius and Side Lengths $\left(\displaystyle \sum_{cycl}(b+c-a)^2\cdot\sum_{cycl}(b+c-a)^3\ge 2592\sqrt{3}r^5\right)$
- An Inequality with Exradii and an Altitude $\left(\displaystyle \sqrt{\frac{1}{r_b^2}+\frac{1}{r_c}+1}+\sqrt{\frac{1}{r_c^2}+\frac{1}{r_b}+1}\ge 2\sqrt{\frac{1}{h_a^2}+\frac{1}{h_a}+1}\right)$
- Leuenberger's Inequality for Medians, Inradius and Circumradius $\left(\displaystyle m_a+m_b+m_c\le 4R+r\right)$
- Adil Abdulayev's Inequality With Angles, Medians, Inradius and Circumradius $\left(\displaystyle \frac{A}{m_a}+\frac{B}{m_b}+\frac{C}{m_c}\le \frac{3\pi}{4R+r}\right)$
- An Inequality with Sides, Cosines, and Semiperimeter $\left(\displaystyle \sum_{cycl}a^2(b\cos B+c\cos C)\le \frac{8s^3}{9}\right)$
- Seyran Ibrahimov's Inequality $\left(\displaystyle \sqrt{3}s\cdot\sum_{cycl}m_a\le 20R^2+r^2\right)$
- An Inequality in Triangle, with Sides and Medians III $\left(\displaystyle \sum_{cycl}\frac{(m_b+m_c-m_a)^3}{m_a}\ge\frac{3}{4}(a^2+b^2+c^2)\right)$
- Leo Giugiuc's Inequality in Triangle, Solely with Cotangents $\left(\cot A +\cot B+\cot C\ge 2\sqrt{2}-1\right)$
- An Inequality in Triangle with Side Lengths and Circumradius $\left(\displaystyle\displaystyle\frac{ab}{\sqrt{a^2+b^2}}+\frac{bc}{\sqrt{b^2+c^2}}+\frac{ca}{\sqrt{c^2+a^2}}\le\frac{3\sqrt{6}R}{2}\right)$
- All Trigonometric Inequality in Triangle $\left(\displaystyle 3\sum_{cycl}\cos A\ge 2\sum_{cycl}\sin A\sin B\right)$
- An Inequality with Two Sets of Cevians $\left(\displaystyle \frac{27[A'B'C']}{[A''B''C'']}\leq \Bigr(\frac{BA'}{BA''}+\frac{CB'}{CB''}+\frac{AC'}{AC''}\Bigr)^3\right)$
- An Inequality with the Most Important Cevians $\left(\displaystyle \frac{m_am_bm_c}{h_ah_bh_c}\ge\frac{\ell_a^2+\ell_b^2+\ell_c^2}{\ell_a\ell_b+\ell_b\ell_c+\ell_c\ell_a}\right)$
- An Inequality in Triangle with a Constraint $(a\sqrt{2}\ge b+c)$
- Problem 11984 From the American Mathematical Monthly $\left(\displaystyle a^6+b^6+c^6\ge 5184\cdot r^6\right)$
- A Long Cyclic Inequality of Degree 4 $\left(\displaystyle 4\cdot\sum_{cycl}ab\cdot\sum_{cycl}a-\left(\sum_{cycl}a\right)^3\ge\frac{\displaystyle 3\sum_{cycl}ab\left[4\sum_{cycl}ab-\left(\sum_{cycl}a\right)^2\right]}{\displaystyle \sum_{cycl}a}\right)$
- An Inequality of Degree 3 with Inradius $\left(\displaystyle \sum_{cycl}(a+b-c)^3+24abc\ge 648\sqrt{3}r^3\right)$
- A Cyclic Inequality from the 6th IMO, 1964 $\left(\displaystyle \sum_{cycl}a^2(b+c-a)\le 3abc\right)$
- An Area Inequality in Right Triangle $\left(\displaystyle \frac{[\Delta ABD]+[\Delta ACE]}{[\Delta ADE]}\ge\sqrt{2}\right)$
- An Angle Inequality in Triangle with Perpendicular Medians $\left(\displaystyle \cos A\ge \frac{4}{5}\right)$
- An Inequality in Triangle form the 1996 APMO $\left(\sqrt{a+b-c}+\sqrt{b+c-a}+\sqrt{c+a-b}\le\sqrt{a}+\sqrt{b}+\sqrt{c}\right)$
- An Inequality in a Nonobtuse Triangle $\left(R\sqrt{2}\le h_a\right)$
- An Inequality in Triangle with Radicals, Semiperimeter and Inradius $\left(\displaystyle \sqrt{\frac{a+b}{s-b}}+ \sqrt{\frac{b+c}{s-c}} +\sqrt{\frac{c+a}{s-a}}\le \frac{\sqrt{a^2+b^2+c^2}}{r}\right)$
- Dan Sitaru's Inequality with Radicals and Cosines $\left(\displaystyle (a^2+b^2+c^2)^3\ge 6^3(abc)^2\cos A\cos B\cos C\right)$
- Lorian Saceanu's Cyclic Inequalities with Three Variables $\left(\displaystyle 2+\sum_{cycl}\frac{a}{b}\ge \sum_{cycl}\frac{a}{c}+2\frac{ab+bc+ca}{a^2+b^2+c^2}\right)$
- An Inequality for the Tangent to the Incircle $\left(\displaystyle DE\le\frac{1}{8}(AB+BC+CA)\right)$
- Lorian Saceanu's Sides And Angles Inequality $\left(\displaystyle \frac{\pi}{3}\le \frac{a\alpha + b\beta + c\gamma}{a+b+c}\le \arccos\left(\frac{r}{R}\right)\right)$
- An Inequality in Triangle with Radicals, Semiperimeter, Incenter and Inradius $\left(\displaystyle \frac{AI+BI+CI}{r}+3\ge\left(\sum_{cycl}\sqrt{s-a}\right)\left(\sum_{cycl}\frac{1}{\sqrt{s-a}}\right)\right)$
- An Inequality in Triangle for Side Lengths, Cycled in Two Ways $\left(\displaystyle 3\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}-1\right)\ge 2\left(\frac{b}{a}+\frac{a}{c}+\frac{c}{b}\right)\right)$
- Lorian Saceanu's Inequality for All Triangles $\left(\displaystyle \sin 2A+\sin 2B+\sin 2C\ge 4\sin 2A\cdot\sin 2B\cdot\sin 2C\right)$
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