Inequality $\sqrt{1+x^2}+\sqrt{1+y^2}+\sqrt{1+z^2} \le \sqrt{2}(x+y+z)$ 
If $x>0$, $y>0$, $z>0$ and $xyz = 1$ then
  $$\sqrt{1+x^2}+\sqrt{1+y^2}+\sqrt{1+z^2} \le \sqrt{2}(x+y+z)$$

I tried using $\displaystyle x = \frac{a}{b},y = \frac{b}{c}$ and $\displaystyle z = \frac{c}{a}$ substitution,
$\displaystyle \sum_{cyc} \frac{\sqrt{a^2+b^2}}{b} \le \sqrt{2}\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right) \\ \displaystyle  \iff 6+ 2\sum_{cyc}\frac{\sqrt{(a^2+b^2)(b^2+c^2)}}{bc} \le \sum_{cyc}\frac{a^2+b^2}{b^2} + 4\sum_{cyc}\frac{b}{a}$
So it would suffice if we showed $\displaystyle \sum_{cyc}\frac{\sqrt{(a^2+b^2)(b^2+c^2)}}{bc} \le 2\sum_{cyc}\frac{b}{a}$
Squaring again does not lead anywhere nice. Is the last inequality true ? If not is there a better way of showing the result ?

To use Lagrange multipliers, we can define 
$$L = \sqrt2(x+y+z) -(\sqrt{x^2+1}+\sqrt{y^2+1}+\sqrt{z^2+1}) - \lambda(xyz-1)$$
Setting the partial derivatives to zero, e.g. $\dfrac{\partial L}{\partial x} = \sqrt2-\dfrac{x}{\sqrt{x^2+1}} - \lambda yz = 0 $ easily gives using the constraint,
$$\lambda = \sqrt2 x - \frac{x^2}{\sqrt{x^2+1}} = \sqrt2 y - \frac{y^2}{\sqrt{y^2+1}}= \sqrt2 x - \frac{z^2}{\sqrt{z^2-1}}$$
$\sqrt2 t - \dfrac{t^2}{\sqrt{t^2 + 1}}$ is strictly increasing $\implies x = y = z = 1$ for the only extremum, which is easily seen to be a minimum, from which the inequality follows.
