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?

You already have a good method, here is an alternative with only AM-GM. First, we have
$xyz = x+y+z \geqslant 3\sqrt[3]{xyz}\implies xyz \geqslant 3\sqrt3$.  Further, again using AM-GM,
$$\sqrt{\frac{t^4}9+\frac{t^4}9+\frac{t^4}9+1}\geqslant \sqrt{\frac4{3\sqrt3}t^3}=\frac2{3^{3/4}}t^{3/2}$$
$\begin{align}
\implies \sqrt{\frac13x^4+1}+\sqrt{\frac13y^4+1}+\sqrt{\frac13z^4+1} &\geqslant \frac2{3^{3/4}}(x^{3/2}+y^{3/2}+z^{3/2}) 
\\ &\geqslant \frac2{3^{3/4}}\cdot 3\,(xyz)^{1/2} 
\\ &\geqslant \frac6{3^{3/4}}\,(3\sqrt3)^{1/2}=6
\end{align}
$
with equality possible iff $x=y=z=\sqrt3$, so we have the minimum.

P.S. Yet another way would be to show the estimate $\sqrt{t^4/3+1} \geqslant \sqrt3 t -1$ holds, and then $\sum \sqrt{x^4/3+1} \geqslant \sqrt3(\sum x)-3 = \sqrt3(xyz)-3  \geqslant \sqrt3\cdot 3\sqrt3-3 = 6$
