$\;32\displaystyle\left(\sum_{c}\frac{1}{7+(x-3)^2}\right)\leq \sum_{c} \frac{x^2+yz}{y+z}+6$

Question

Let $x, y, z > 0$

Prove that $\;32\displaystyle\left(\displaystyle\sum_{c}\frac{1}{7+(x-3)^2}\right)\leq \displaystyle\sum_{c} \frac{x^2+yz}{y+z}+6$

My work,

$\displaystyle\sum_{c} \frac{x^2+2yz}{y+z} = \displaystyle\sum_{c} \frac{x^2+yz}{y+z} + \displaystyle\sum_{c} \frac{x^2+yz}{y+z}$

$\leq \displaystyle\sum_{c} \frac{x^2+yz}{y+z} + \displaystyle\sum_{c} \frac{yz}{2 \sqrt{yz}} = \displaystyle\sum_{c} \frac{x^2+yz}{y+z} + \displaystyle\sum_{c} \frac{\sqrt{yz}}{2}$

$\leq \displaystyle\sum_{c} \frac{x^2+yz}{y+z} + \frac{1}{2}\displaystyle\sum_{c}y$ ---[1]

And

$\displaystyle\sum_{c}\frac{x^2+2yz}{y+z}-\frac{3}{2}(x+y+z)=\frac{1}{2}\frac{\displaystyle\sum_{c}((-y-z+x)^2+yz)(y-z)^2)}{(y+z)(x+z)(x+y)} \geq 0$ ---[2]

From [1], [2], $\displaystyle\sum_{c} \frac{x^2+yz}{y+z} \geq \displaystyle\sum_{c}x$

Please suggest how to do to get the LHS. Thank you.

Answer 1

By your work it's enough to prove that $$32\sum_{cyc}\frac{1}{x^2-6x+16}\leq x+y+z+6$$ or $$\sum_{cyc}\left(x+2-\frac{32}{x^2-6x+16}\right)$$ or $$\sum_{cyc}\frac{x(x-2)^2}{x^2-6x+16}\geq0.$$ Also, we have $$\sum_{cyc}\frac{x^2+yz}{y+z}-x-y-z=\sum_{cyc}\left(\frac{x^2+yz}{y+z}-\frac{y+z}{2}\right)=$$ $$=\sum_{cyc}\frac{2x^2-y^2-z^2}{2(y+z)}=\sum_{cyc}(x^2-y^2)\left(\frac{1}{2(y+z)}-\frac{1}{2(x+z)}\right)=$$ $$=\sum_{cyc}\frac{(x-y)^2(x+y)}{2(y+z)(x+z)}\geq0.$$