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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.

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