<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] 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:
$$\sum_{cyc}\frac{(a+b)^2}{a^2+b^2+2c^2}\leq\sum_{cyc}\left(\frac{a^2}{a^2+c^2}+\frac{b^2}{b^2+c^2}\right)=\sum_{cyc}\left(\frac{a^2}{a^2+c^2}+\frac{c^2}{c^2+a^2}\right)=3.$$
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