### Solution

If $x=z,\,$ then $x=y=z\,$ and there is nothing to proof. So assume $x\lt z.\,$ Introduce $t=\displaystyle\sqrt{\frac{y}{x}}\,$ and $k=\displaystyle\sqrt{\frac{z}{x}}.\,$ Obviously, $1\le t\le k,\,$ and we have to prove that

$\displaystyle \frac{(1+k)^2}{12k}\cdot\frac{(t+1+k)^2}{t^2+k^2+1}\ge 1.$

Consider the function $\displaystyle f(t)=\frac{(1+k)^2}{12k}\cdot\frac{(t+1+k)^2}{t^2+k^2+1}\,$ for $t\in [1,k].$

$\displaystyle f'(t)=\frac{(1+k)^2}{6k}\cdot\frac{[1+k^2-(1+k)t](t+1+k)}{(t^2+k^2+1)^2}.$

The only critical point of $f\,$ is $\displaystyle t'=\frac{1+k^2}{1+k}\,$ and $1\lt t'\lt k,\,$ implying that $\min f\in \{f(1),f(k)\}.\,$ But $f(1)\gt 1\,$ and $f(k)\gt 1,\,$ which completes the proof.

### Acknowledgment

The unusual problem by Lorian Saceanu has been kindly communicated to me by Leo Giugiuc, along with his solution.