Inequalities in Triangle

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.)

1. $\sin\frac{A}{2}\le\frac{a}{2\sqrt{bc}}$

2. $\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}\le\frac{1}{8}$

3. $\cos (A)+\cos (B)+\cos (C)\le\frac{3}{2}$

4. $m^{2}_{a}+m^{2}_{b}+m^{2}_{c}\le\frac{27}{4}R^{2}$

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

6. $\sin (A)+\sin (B)+\sin (C)\le\frac{3\sqrt{3}}{2}$

7. $\sin (A)\sin (B)\sin (C)\le\frac{3\sqrt{3}}{8}$

8. $a^3b+b^3c+c^3a-a^2b^2-b^2c^2-c^2a^2\ge 0$

9. $m_a l_a+m_b l_b+m_c l_c\ge p^{2}$

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 $a^{2}\ge 4bc\,\sin^{2}(\frac{A}{2}),$ $A\ge 2\sqrt{bc}\sin\frac{A}{2}.$

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:

$\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}\le\frac{abc}{8\sqrt{a^{2}b^{2}c^{2}}}=\frac{1}{8}.$

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

By Lagrange's theorem of moments, if $G$ is the centroid of $\Delta ABC$ and $O$ is the circumcenter,

$\begin{align} 3R^{2} &= OA^{2}+OB^{2}+OC^{2}\\ &= GA^{2}+GB^{2}+GC^{2}+3GO^{2}\\ &\ge GA^{2}+GB^{2}+GC^{2}\\ &=\frac{4}{9}(m^{2}_{a}+m^{2}_{b}+m^{2}_{c}). \end{align}$

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.

Since $a=2R\,\sin (A),$ etc., this is equivalent to the previous inequality.

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.

I have placed the proof of the above in a separate file.

I have placed the proof of the above in a separate file.

The books Problems in Planimetry (especially the second volume) by V. V. Prasolov (Nauka, Moscow, 1986 (Russian)) are an excellent starting point.