How to approach this minimization problem? I see this problem somewhere (it is from an olympiad or something but Im not sure where it comes):

Minimize $$\frac1{\sqrt{y+\frac1x+\frac12}}+\frac1{\sqrt{z+\frac1y+\frac12}}+\frac1{\sqrt{x+\frac1z+\frac12}}\tag1$$ for $x,y,z>0$ and $xyz=1$.

My first thought was approach this problem trying to find the points where $\nabla f(x,y)=0$ and $\partial^2 f(x,y)$ is positive definite for $f$ being the function defined by $(1)$ and $xy=\frac1z$, that is, I wanted to find the local minima for such $f$ and after choose the absolute minimum.
However this approach is messy, so I guess it must be a better way to approach this problem. So my question is, do you know a better way to approach this problem? I dont have a clue about how to handle it.
 A: First, take $\displaystyle (x, y, z) = \left(t, t, \frac{1}{t^2}\right)$ and make $t \to +\infty$ to see that the infimum is no greater than $\sqrt{2}$. Next it will be proved that the infimum is $\sqrt{2}$.
Set $x = a^3$, $y = b^3$, $z = c^3$, then $abc = 1$ and\begin{align*}
(1) &= \frac{1}{\sqrt{\frac{b^3}{abc} + \frac{abc}{a^3} + \frac{1}{2}}} + \frac{1}{\sqrt{\frac{c^3}{abc} + \frac{abc}{b^3} + \frac{1}{2}}} + \frac{1}{\sqrt{\frac{a^3}{abc} + \frac{abc}{c^3} + \frac{1}{2}}}\\
&= \sqrt{\frac{ca^2}{ab^2 + bc^2 + \frac{1}{2} ca^2}} + \sqrt{\frac{ab^2}{bc^2 + ca^2 + \frac{1}{2} ab^2}} + \sqrt{\frac{bc^2}{ca^2 + ab^2 + \frac{1}{2} bc^2}}.
\end{align*}
Denote $u = bc^2$, $v = ca^2$, $w = ab^2$, then it suffices to prove$$
\sum_{\mathrm{cyc}} \sqrt{\frac{u}{\frac{u}{2} + v + w}} > \sqrt{2} \tag{2}
$$
for all $u, v, w > 0$, where $\sum\limits_{\mathrm{cyc}}$ means cyclic summation. Without loss of generality, assume that $u \geqslant v \geqslant w$. Since\begin{align*}
&\mathrel{\phantom{\Longleftrightarrow}} \frac{v}{\frac{v}{2} + w + u} + \frac{w}{\frac{w}{2} + u + v} > \frac{v}{\frac{v}{2} + u}\\
&\Longleftrightarrow \frac{w}{\frac{w}{2} + u + v} > \frac{v}{\frac{v}{2} + u} - \frac{v}{\frac{v}{2} + w + u} = \frac{vw}{\left(\frac{v}{2} + u\right) \left(\frac{v}{2} + u + w\right)}\\
&\Longleftrightarrow \left(\frac{v}{2} + u\right) \left(\frac{v}{2} + u + w\right) > v \left(\frac{w}{2} + u + v\right)\\
&\Longleftrightarrow u^2 - \frac{3}{4} v^2 + uw > 0,
\end{align*}
then\begin{align*}
\sum_{\mathrm{cyc}} \sqrt{\frac{u}{\frac{u}{2} + v + w}} &= \sqrt{\frac{u}{\frac{u}{2} + v + w}} + \sqrt{\frac{v}{\frac{v}{2} + w + u}} + \sqrt{\frac{w}{\frac{w}{2} + u + v}}\\
&> \sqrt{\frac{u}{\frac{u}{2} + v + w}} + \sqrt{\frac{v}{\frac{v}{2} + w + u} + \frac{w}{\frac{w}{2} + u + v}}\\
&> \sqrt{\frac{u}{\frac{u}{2} + v + w}} + \sqrt{\frac{v}{\frac{v}{2} + u}} \geqslant \sqrt{\frac{u}{\frac{u}{2} + 2v}} + \sqrt{\frac{v}{\frac{v}{2} + u}}\\
&= \sqrt{\frac{1}{\frac{1}{2} + 2t}} + \sqrt{\frac{t}{\frac{t}{2} + 1}} = \sqrt{\frac{2}{4t + 1}} + \sqrt{\frac{2t}{t + 2}},
\end{align*}
where $\displaystyle t = \frac{v}{u} \in (0, 1]$. Because\begin{align*}
\sqrt{\frac{2}{4t + 1}} + \sqrt{\frac{2t}{t + 2}} > \sqrt{2} &\Longleftrightarrow \sqrt{\frac{2}{4t + 1}} > \sqrt{2} - \sqrt{\frac{2t}{t + 2}}\\
&\Longleftrightarrow \frac{2}{4t + 1} > 2 \left(1 + \frac{t}{t + 2} - 2 \sqrt{\frac{t}{t + 2}}\right)\\
&\Longleftrightarrow 2 \sqrt{\frac{t}{t + 2}} > 1 + \frac{t}{t + 2} - \frac{1}{4t + 1} = \frac{8t (t + 1)}{(t + 2)(4t + 1)}\\
&\Longleftrightarrow 1 > \frac{4\sqrt{t} (t + 1)}{\sqrt{t + 2}(4t + 1)}\\
&\Longleftrightarrow (t + 2)(4t + 1)^2 > 16t(t + 1)^2\\
&\Longleftrightarrow 8t^2 + t + 2 > 0,
\end{align*}
then (2) holds. Therefore,$$
\inf_{\substack{x, y, z > 0\\xyz = 1}} \left\{\frac{1}{\sqrt{y + \frac{1}{x} + \frac{1}{2}}} + \frac{1}{\sqrt{z + \frac{1}{y} + \frac{1}{2}}} + \frac{1}{\sqrt{x + \frac{1}{z} + \frac{1}{2}}}\right\} = \sqrt{2}.
$$
