Prove Inequality for $x,y,z$ Positive Real Numbers

[anemone]

Given $x,\,y,\,z$ are positive real numbers. Prove that

$\dfrac{xy}{x^2+xy+y^2}-\dfrac{1}{9}+\dfrac{yz}{y^2+yz+z^2}-\dfrac{1}{9}+\dfrac{zx}{z^2+zx+x^2}-\dfrac{1}{9}\le \dfrac{2\sqrt{xy+yz+zx}}{3\sqrt{x^2+y^2+z^2}}$

 

[anemone]

Solution of other:

Spoiler

Note that

$$\begin{align*}\sum_{cyclic}^{}\dfrac{xy}{x^2+xy+y^2}&=1-\left(\sum_{cyclic}^{}\dfrac{1}{3}-\dfrac{xy}{x^2+xy+y^2}\right)\\&=1-\sum_{cyclic}^{}\dfrac{(x-y)^2}{3(x^2+xy+y^2)}\\&\le 1-\sum_{cyclic}^{}\dfrac{(x-y)^2}{3(x^2+xy+y^2+z^2+yz+zx)}\\&=1-\dfrac{2}{3}\cdot \dfrac{x^2+y^2+z^2-xy-yz-zx}{x^2+y^2+z^2+xy+yz+zx}\\&=\dfrac{1}{3}+\dfrac{2}{3}\cdot \dfrac{2(xy+yz+zx)}{x^2+y^2+z^2+xy+yz+zx}\\&\le \dfrac{1}{3}+\dfrac{2}{3}\cdot \dfrac{2(xy+yz+zx)}{\sqrt{x^2+y^2+z^2}\sqrt{xy+yz+zx}}\\&=\dfrac{1}{3}+\dfrac{2}{3}\sqrt{\dfrac{xy+yz+zx}{x^2+y^2+z^2}} \end{align*}$$

Therefore we get

$\dfrac{xy}{x^2+xy+y^2}-\dfrac{1}{9}+\dfrac{yz}{y^2+yz+z^2}-\dfrac{1}{9}+\dfrac{zx}{z^2+zx+x^2}-\dfrac{1}{9}\le \dfrac{2\sqrt{xy+yz+zx}}{3\sqrt{x^2+y^2+z^2}}$ and we're hence done.