<s>[INST] How prove this inequality with $x+y+z=1$ 
let $x,y,z>0$,and such $$x+y+z=1$$
show that:$$\dfrac{(1+xy+yz+xz)(1+3x^2+3y^2+3z^2)}{9(x+y)(y+z)(x+z)}
\ge \left(\dfrac{x\sqrt{1+x}}{\sqrt[4]{3+9x^2}}+\dfrac{y\sqrt{1+y}}{\sqrt[4]{3+9y^2}}
+\dfrac{z\sqrt{1+z}}{\sqrt[4]{3+9z^2}}\right)^2$$

My idea: maybe we can prove $LHS\ge 1,RHS\le\dfrac{2}{3}?$
(because when $x=y=z=\dfrac{1}{3},LHS=1,RHS=\dfrac{2}{3}$)
or 
$$(1+xy+yz+xz)[1+3(x^2+y^2+z^2)]\ge 9(x+y)(y+z)(x+z)$$
and  $$\left(\dfrac{x\sqrt{1+x}}{\sqrt[4]{3+9x^2}}+\dfrac{y\sqrt{1+y}}{\sqrt[4]{3+9y^2}}
+\dfrac{z\sqrt{1+z}}{\sqrt[4]{3+9z^2}}\right)^2\le\dfrac{2}{3}?$$
can you use computer test it?
I know this well know inequality
$$9(x+y)(y+z)(x+z)\ge 8(x+y+z)(xy+yz+xz)=8(xy+yz+xz)$$
 [/INST] We can show LHS $\ge 1 \iff $
$$ (1+xy+yz+xz)[1+3(x^2+y^2+z^2)]\ge 9(x+y)(y+z)(x+z) $$
Let $p = x+y+z = 1, q = xy+yz+zx \le \frac13, r = xyz$.  Then we have the inequality as
$$(1+q)[1+3(p^2-2q)] \ge 9(pq-r) \iff 4+9r \ge 11q + 6q^2$$
By Schur $p^3+9r \ge 4pq \implies 1+9r \ge 4q$, so it is enough to show that $3 \ge 7q + 6q^2$ which follows from $q \le \frac13$. 
For the second part, i.e. RHS $\le \frac23$, it is sufficient to note that
$$ \frac{x\sqrt{1+x}}{\sqrt[4]{3+9x^2}} \le \sqrt{\frac23}x \iff (3x-1)^2 \ge 0$$
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