Find the minimum value of the expression $\sqrt{\frac{1}{3}x^4+1}+\sqrt{\frac{1}{3}y^4+1}+\sqrt{\frac{1}{3}z^4+1}$ Let $x, y, z$ be positive real numbers such that $x + y + z = xyz$.
Find the minimum value of the expression $$\sqrt{\frac{1}{3}x^4+1}+\sqrt{\frac{1}{3}y^4+1}+\sqrt{\frac{1}{3}z^4+1}$$
My attempt:
By using Cauchy-Schwarz inequality and AM-GM inequality,
$$\sqrt{\frac{1}{3}t^4+1}\cdot\sqrt{3+1}\geq t^2+1$$
$$(x^2+y^2+z^2)(x+y+z)\geq3(xyz)^{\frac{2}{3}}\cdot 3(xyz)^{\frac{1}{3}}=9xyz$$
$$\therefore \sqrt{\frac{1}{3}x^4+1}+\sqrt{\frac{1}{3}y^4+1}+\sqrt{\frac{1}{3}z^4+1}\geq \frac{1}{2}(x^2+y^2+z^2)+\frac{3}{2}$$
$$\geq\frac{1}{2}\left (\frac{9xyz}{x+y+z}+\frac{3}{2} \right)=6$$
with equality if and only if $x=y=z=\sqrt{3}$
What are the alternative methods to solve this question?

Your method is very straightforward. I am going just to post another approach that may contain a new idea.
Firstly;
$$xyz=x+y+z\ge 3 \sqrt[3] {xyz}\implies (xyz)^3\ge 27xyz\implies x+y+z=xyz\ge 3\sqrt3.$$
Secondly, a simple computation shows that $f(t)=\sqrt{\frac {t^4}{3}+1}$ is a convex function. Therefore, by Jensen's inequality, we get:
$$\sqrt{\frac {x^4}{3}+1}+\sqrt{\frac {y^4}{3}+1}+\sqrt{\frac {z^4}{3}+1}\ge 3\sqrt {\frac {(\frac {x+y+z}{3})^4}{3}+1}\ge 3\sqrt {\frac {(\frac {3\sqrt 3}{3})^4}{3}+1}=6. $$
More information about Jensen's inequality is available here.
