Let $f(x)=\sqrt{2}x-\sqrt{x^2+1}-\dfrac{\sqrt{2}\ln x}{2}$, note that (from graphical method) for $0<x<1$, we have the function $g(x)=\sqrt{2}x-\sqrt{x^2+1}$ lies above the function $h(x)=\dfrac{\sqrt{2}\ln x}{2}$ so $\sqrt{2}x-\sqrt{x^2+1}\ge\dfrac{\sqrt{2}\ln x}{2}$ holds for $0<x\le 1$, equality occurs at $x=1$.
Next, if we could prove $\sqrt{2}x-\sqrt{x^2+1}>\dfrac{\sqrt{2}\ln x}{2}$ for $x>1$, then the result will follow.
By using differentiation method, we find
$f'(x)=\sqrt{2}-\dfrac{x}{\sqrt{x^2+1}}-\dfrac{\sqrt{2}}{2x}=\dfrac{(2x-1)\sqrt{x^2+1}-\sqrt{2}x^2}{x\sqrt{2(x^2+1)}}$,
$f''(x)=\dfrac{1}{\sqrt{2}x^2}-\dfrac{1}{(x^2+1)\sqrt{x^2+1}}$ and $f'(x)=0$ iff $P(x)=(2x-1)\sqrt{x^2+1}-\sqrt{2}x^2=0$.
It's not hard to see $P(1)=0$, and $f''(1)>0$ so $(1,\,0)$ is a minimum point.
Note that $(2x-1)\sqrt{x^2+1}-\sqrt{2}x^2\div (x-1)$ gives $(x-1)(2x^3-2x^2+3x-1)=0$ and since $Q(x)=2x^3-2x^2+3x-1$ is a strictly increasing function beyond $x=1$ we can therefore conclude that $f(x)=\sqrt{2}x-\sqrt{x^2+1}-\dfrac{\sqrt{2}\ln x}{2}\ge 0$ for $x>0$.
Now, for $x,\,y,\,z>0$ and $xyz=1$, we can make the three inequalities as shown below and adding them up gives the desired result.
$\sqrt{2}x-\sqrt{x^2+1}\ge \dfrac{\sqrt{2}\ln x}{2}$
$\sqrt{2}y-\sqrt{y^2+1}\ge \dfrac{\sqrt{2}\ln y}{2}$
$\sqrt{2}z-\sqrt{z^2+1}\ge \dfrac{\sqrt{2}\ln z}{2}$
$\therefore \sqrt{x^2+1}+\sqrt{y^2+1}+\sqrt{z^2+1}\le\sqrt{2}(x+y+z)$.
