<s>[INST] How prove that $xyz+\sqrt{x^2y^2+y^2z^2+x^2z^2}\ge \frac{4}{3}\sqrt{xyz(x+y+z)}$ let $x,y,z>0$,and such that
$x^2+y^2+z^2=1$,prove that
$$xyz+\sqrt{x^2y^2+y^2z^2+x^2z^2}\ge \dfrac{4}{3}\sqrt{xyz(x+y+z)}$$
Does this have a nice solution? Thank you everyone.
 [/INST] We need to prove that
$$xyz+\sqrt{(x^2+y^2+z^2)(x^2y^2+x^2z^2+y^2z^2)}\geq\frac{4}{3}\sqrt{xyz(x+y+z)(x^2+y^2+z^2)}.$$
Let $x+y+z=3u$, $xy+xz+yz=3v^2$ and $xyz=w^3$.
Hence, we need to prove that $f(w^3)\geq0,$ where
$$f(w^3)=w^3+3\sqrt{(3u^2-2v^2)(3v^4-2uw^3)}-4\sqrt{u(3u^2-2v^2)w^3}.$$
But by AM-GM
$$f'(w^3)=1-\frac{3u\sqrt{3u^2-2v^2}}{\sqrt{3v^4-2uw^3}}-2\sqrt{\frac{u(3u^2-2v^2)}{w^3}}\leq1-\sqrt{\frac{u(3u^2-2v^2)}{w^3}}\leq0,$$
which says that $f$ is a decreasing function and it's enough to prove our inequality 
for a maximal value of $w^3$, which happens for equality case of two variables.
Since our inequality is homogeneous, it's enough to do it for $y=z=1$, which gives
$$x+\sqrt{(x^2+2)(2x^2+1)}\geq\frac{4}{3}\sqrt{x(x+2)(x^2+2)}$$ or after squaring of the both sides
$$x^4-16x^3+11x^2-32x+9+9x\sqrt{(x^2+2)(2x^2+1)}\geq0.$$
Now, by C-S $$\sqrt{(x^2+2)(2x^2+1)}=\sqrt{(x^2+1+1)(x^2+x^2+1)}\geq x^2+x+1.$$
Id est, it remains to prove that
$$x^4-16x^3+11x^2-32x+9+9x(x^2+x+1)\geq0$$ or
$$(x-1)^2(x^2-5x+9)\geq0.$$
Done!
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