Home » Uncategorized » Inequality 50 (Vasile Cirtoaje)

# Inequality 50 (Vasile Cirtoaje)

Problem:

For all $\displaystyle a,b,c>0$ prove that

$\displaystyle \frac{3a^2-2ab-b^2}{a^2+b^2}+\frac{3b^2-2bc-c^2}{b^2+c^2}+\frac{3c^2-2ca-a^2}{c^2+a^2}\geq 0$.

Solution:

It holds that $\displaystyle \frac{3a^2-2ab-b^2}{a^2+b^2}=\frac{2(a^2-b^2)+(a-b)^2}{a^2+b^2}$. Thus, we only need to prove that

$\displaystyle \sum\frac{(a-b)^2}{a^2+b^2}\geq -2\sum\frac{a^2-b^2}{a^2+b^2}$.

Observe that

$\displaystyle \sum\frac{a^2-b^2}{a^2+b^2}=-\frac{(a^2-b^2)(b^2-c^2)(c^2-a^2)}{(a^2+b^2)(b^2+c^2)(c^2+a^2)}$.

From the above result we have to prove now that

$\displaystyle \sum\frac{(a-b)^2}{a^2+b^2}\geq 2\frac{(a^2-b^2)(b^2-c^2)(c^2-a^2)}{(a^2+b^2)(b^2+c^2)(c^2+a^2)}$.

Using the AM – GM Inequality we get that

$\displaystyle \sum\frac{(a-b)^2}{a^2+b^2}\geq 3\sqrt[3]{\frac{(a-b)^2(b-c)^2(c-a)^2}{(a^2+b^2)(b^2+c^2)(c^2+a^2)}}$.

So, it is enough to prove that

$\displaystyle 3\sqrt[3]{\frac{(a-b)^2(b-c)^2(c-a)^2}{(a^2+b^2)(b^2+c^2)(c^2+a^2)}}\geq 2\frac{(a^2-b^2)(b^2-c^2)(c^2-a^2)}{(a^2+b^2)(b^2+c^2)(c^2+a^2)}$.

The last Inequality can be reduced to the $\displaystyle 27(a^2+b^2)^2(b^2+c^2)^2(c^2+a^2)^2\geq 8(a-b)(b-c)(c-a)(a+b)^3(b+c)^3(c+a)^3$.

So, it is enough to check the Inequality

$\displaystyle 3(a^2+b^2)^2\geq 2|a-b|(a+b)^3$.

This one can be prove by the following way. Set $\displaystyle a^2+b^2=x$ and $\displaystyle a^2-b^2=y$, with $\displaystyle x\geq y$ and remake the Inequality to the form

$\displaystyle 3\frac{(a^2+b^2)^2}{(a+b)^2}\geq 2|a^2-b^2|$.

Then, the Inequality substitutes to

$\displaystyle 3\frac{x^2}{x+\sqrt{x^2-y^2}}\geq 2y$.

But this one holds because we have that

$\displaystyle 2xy++2y\sqrt{x^2-y^2}\leq 2xy+(x^2-y^2)+y^2=x^2+2xy\leq 3x^2$,

due to the AM – GM Inequality and the hypothesis $\displaystyle x \geq y$, Q.E.D.

P.S The following nice Inequality also holds:

$\displaystyle \sum \frac{3a^2-2ab-b^2}{3a^2+2ab+3b^2}\geq 0$.

This Inequality belongs to Thomas Mildorf and the proof for this Inequality is the same as the above.