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.

Link download http://tuan.nguyentrung.googlepages.com/AMM_11306.pdf

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