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 $A=B=C=60^{\circ}.$

$\displaystyle\sin\frac{A}{2}+\sin\frac{B}{2}+\sin\frac{C}{2}\le\frac{3}{2}.$

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.

$a+b+c\le 3\sqrt{3}R$

We know that

$\begin{align} 4m^{2}_{a}=2(b^{2}+c^{2})-a^{2},\\ 4m^{2}_{b}=2(c^{2}+a^{2})-b^{2},\\ 4m^{2}_{c}=2(a^{2}+b^{2})-c^{2},\\ \end{align}$

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.

$\displaystyle\tan\frac{A}{2}+\tan\frac{B}{2}+\tan\frac{C}{2}\ge\sqrt{3}$

We know that

$\displaystyle\tan\frac{A}{2}\tan\frac{B}{2}+\tan\frac{B}{2}\tan\frac{C}{2}+\tan\frac{C}{2}\tan\frac{A}{2}=1.$

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.

 

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