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# Inequality 36(Christos Patilas)

Problem:

If $\displaystyle a,b,c$ are side lengths of a triangle, and $\displaystyle E,R$ be its area and circumradius respectively, satisfying the equality $\displaystyle a+b+c=3$, prove that

$\displaystyle \frac{3}{a^2+b^2+c^2}\geq 2^{\sqrt{2}}\cdot \left(ER\right)^{\frac{1}{\sqrt{2}}}$.

Solution:

Let $\displaystyle E=\frac{abc}{4R}$.

Then the given inequality is of the form

$\displaystyle \frac{3}{a^2+b^2+c^2}\geq 2^{\sqrt{2}}\cdot \left(\frac{abc}{4}\right)^{\frac{1}{\sqrt{2}}}$.

Removing the square roots of $\displaystyle 2$ we get that

$\displaystyle \left(\frac{3}{a^2+b^2+c^2}\right)^{\sqrt{2}}\geq abc$.

Let us now introduce the $\displaystyle u,v^2,w^3$, that is

$\displaystyle3u=a+b+c,3v^2=ab+bc+ca,w^3=abc$.

Then the inequality takes the form

$\displaystyle w^3\leq u^3\left(\frac{u^2}{3u^2-2v^2}\right)^{\sqrt{2}}$.

Thus $\displaystyle w^3$ attains its maximum when $\displaystyle 2$ of the $\displaystyle \left\{a,b,c\right\}$ are equal.

So, we need only need to prove the inequality for $\displaystyle a=b$, or we need to prove that for $\displaystyle 2a+c=1$ it holds that

$\displaystyle \left(\frac{3}{2a^2+c^2}\right)^{\sqrt{2}}\geq a^2c$,

which is true, Q.E.D.