<s>[INST] Contradiction on an Inequality 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!
 [/INST] 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 $$c^2-(a+b)c+a^2-ab+b^2\geq c^2-(a+b)c+\frac{1}{4}(a+b)^2=\frac{1}{4}(2c-a-b)^2\geq0.$$
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