# Solution of problem 11306 in AMM

Let $a,b,$ and $c$ be the lengths of the sides of a nondegenerate triangle, let $p=(a+b+c)/2$, and let $r$ and $R$ be the inradius and circumradius of the triangle, respectively. Show that $\dfrac{a}{2}\cdot \dfrac{4r-R}{R}\leq\sqrt{(p-b)(p-c)}\leq \dfrac{a}{2},$

and determine the cases of equality.

My solution.

A)Proof of the 1st inequality:

Assume that $S$ is area of the triangle, then $S=\dfrac{abc}{4R}$ and $S=pr$, therefore that inequality is equivalent to

$\dfrac{a}{2}\cdot\left(\dfrac{4r}{R}-1\right)\leq \sqrt{(p-b)(p-c)}$

$\Leftrightarrow \dfrac{a}{2}\cdot\left(\dfrac{16(p-a)(p-b)(p-c)}{abc}-1\right)\leq \sqrt{(p-b)(p-c)}$

Setting $p-a=x,p-b=y,p-c=z$, then $x>0,y>0,z>0,a=y+z,b=z+x,c=x+y$ and we need only prove $\dfrac{y+z}{2}\cdot\left(\dfrac{16xyz}{(x+y)(y+z)(z+x)}-1\right)\leq \sqrt{yz}$

$\Leftrightarrow \dfrac{8xyz}{(x+y)(z+x)}\leq \sqrt{yz}+\dfrac{y+z}{2}$

$\Leftrightarrow 16xyz\leq (x+y)(z+x)(\sqrt{y}+\sqrt{z})^2 (*)$

By AM-GM we have $x+y\geq 2\sqrt{xy},z+x\geq 2\sqrt{zx}$ and $(\sqrt{y}+\sqrt{z})^2\geq 4\sqrt{yz}$ therefore $(*)$ and also 1st inequality proved.

Equality occur iff $x=y=z$ iff triangle is equilateral.

B)Proof of the 2nd inequality:

By AM-GM we have $\sqrt{(p-b)(p-c)}\leq\dfrac{(p-b)+(p-c)}{2}=\dfrac{a}{2}$

and we’re done. Equality occur iff $p-b=p-c$ iff $b=c$.