How prove this inequality $\sum\limits_{cyc}\frac{x^2}{(2x+y)(2x+z)}\le\frac{1}{3}$ let $x,y,z>0$,show that
$$\dfrac{x^2}{(2x+y)(2x+z)}+\dfrac{y^2}{(2y+z)(2y+x)}+\dfrac{z^2}{(2z+x)(2z+y)}\le\dfrac{1}{3}$$
My try: $$\Longleftrightarrow\sum_{cyc}\dfrac{4x^2}{4x^2+2x(y+z)+yz}\le\dfrac{4}{3}$$
$$\Longleftrightarrow\sum_{cyc}\left(1-\dfrac{2x(y+z)+yz}{4x^2+2x(y+z)+yz}\right)\le\dfrac{4}{3} $$
$$\Longleftrightarrow \sum_{cyc}\dfrac{2x(y+z)+yz}{4x^2+2x(y+z)+yz}\ge\dfrac{5}{3}$$
then I can't.Thank you
 A: I have consider nice methos: use Cauchy-Schwarz inequality
\begin{align*}
\dfrac{a^2}{(2a+b)(2a+c)}&=\dfrac{a^2}{4a^2+2ab+2ac+bc}=\dfrac{a^2}{(2a^2+bc)+2a(a+b+c)}\\
&\le\dfrac{a^2}{9}\left(\dfrac{1}{2a^2+bc}+\dfrac{2}{a(a+b+c)}\right)\\
&=\dfrac{1}{9}\left(\dfrac{2a}{a+b+c}+\dfrac{a^2}{2a^2+bc}\right)
\end{align*}
so
$$\sum_{cyc}\dfrac{a^2}{(2a+b)(2a+c)}\le\dfrac{1}{9}\left(2+\sum\dfrac{a^2}{2a^2+bc}\right)$$
it suffices to prove that
$$\dfrac{a^2}{2a^2+bc}+\dfrac{b^2}{2b^2+ca}+\dfrac{c^2}{2c^2+ab}\le 1$$
which is equivalent to
$$\dfrac{bc}{bc+2a^2}+\dfrac{ca}{ca+2b^2}+\dfrac{ab}{ab+2c^2}\ge 1$$
Using the AM-GM inequality,we hve
$$\dfrac{bc}{bc+2a^2}=\dfrac{b^2c^2}{b^2c^2+2a^2bc}\ge\dfrac{b^2c^2}{b^2c^2+a^2(b^2+c^2)}=\dfrac{b^2c^2}{b^2c^2+a^2c^2+a^2b^2}$$
A: Let $x\geq y\ge z$.
Hence,
$$\frac{1}{3}-\sum_{cyc}\frac{x^2}{(2x+y)(2x+z)}=\sum_{cyc}\left(\frac{x}{3(x+y+z)}-\frac{x^2}{(2x+y)(2x+z)}\right)=$$
$$=\sum_{cyc}\frac{x(x-y)(x-z)}{3(x+y+z)(2x+y)(2x+z)}\geq$$
$$\geq\frac{x(x-y)(x-z)}{3(x+y+z)(2x+y)(2x+z)}+\frac{y(y-x)(y-z)}{3(x+y+z)(2y+x)(2y+z)}=$$
$$=\frac{(x-y)^2(2xy(x+y)+(x^2+5xy+y^2)z-(x+y)z^2)}{3(x+y+z)(2x+y)(2x+z)(2y+x)(2y+z)}\geq0.$$
Done!
