Contradiction on an Inequality

Question

The problem on which I have a problem is this-

Let a, b, c be non-negative real numbers. Prove that $$ \sum_{cyc} {a^2-bc \over 2a^2+b^2+c^2} \ge 0 $$

While solving and after some resolution, we get $$ \sum_{cyc} {(a+b)^2 \over a^2+b^2+2c^2} \le 3 $$ And by C-S, we have, $$ \sum_{cyc} {(a+b)^2 \over a^2+b^2+2c^2} \le \sum_{cyc} {2(a^2+b^2) \over a^2+b^2+2c^2} $$ It rests to prove that $$ \sum_{cyc} {a^2+b^2 \over a^2+b^2+2c^2} \le {3 \over 2} $$ By cyclic substitutions of $x$ for $b^2+c^2$, we get, $$ \sum_{cyc} {a^2+b^2 \over a^2+b^2+2c^2} = \sum_{cyc} {(a^2+b^2) \over (c^2+a^2)+(b^2+c^2)} = \sum_{cyc} {z \over x+y} \le {3 \over 2} $$ But by the Nesbitt's Inequality, $$ \sum_{cyc} {z \over x+y} \ge {3 \over 2} !$$ Can anybody explain me where's the mistake and the correction?
Thanks!

Answer 1

Another solution.

By your work we need to prove that: $$\sum_{cyc}\frac{(a+b)^2}{a^2+b^2+2c^2}\leq3,$$ which is true by C-S: $$\sum_{cyc}\frac{(a+b)^2}{a^2+b^2+2c^2}\leq\sum_{cyc}\left(\frac{a^2}{a^2+c^2}+\frac{b^2}{b^2+c^2}\right)=\sum_{cyc}\left(\frac{a^2}{a^2+c^2}+\frac{c^2}{c^2+a^2}\right)=3.$$

Answer 2

Your first step was not so strong, which gave a wrong inequality.

SOS helps here:

We need to prove that $$\sum_{cyc}\frac{(a-b)(a+c)-(c-a)(a+b)}{2a^2+b^2+c^2}\geq0$$ or $$\sum_{cyc}(a-b)\left(\frac{a+c}{2a^2+b^2+c^2}-\frac{b+c}{2b^2+a^2+c^2}\right)\geq0$$ or $$\sum_{cyc}(a-b)^2(c^2-(a+b)c+a^2-ab+b^2)(2c^2+a^2+b^2)\geq0,$$ which is true because $$c^2-(a+b)c+a^2-ab+b^2\geq c^2-(a+b)c+\frac{1}{4}(a+b)^2=\frac{1}{4}(2c-a-b)^2\geq0.$$