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 ?

WLOG replace the variables by their cubes, while still $xyz=1$:
$$\sum_\text{cyc}\sqrt{1+x^6}\le \sqrt2\sum_\text{cyc}x^3$$
We can homogenize and use the Cauchy-Schwarz inequality:
$$\biggl(\sum_\text{cyc}\sqrt{x^2}\sqrt{y^2z^2+x^4}\biggr)^{\!2}\le\Bigl(\sum_\text{cyc}x^2\Bigr)\Bigl(\sum_\text{cyc}(y^2z^2+x^4)\Bigr)\implies\\ \sum_\text{cyc}\sqrt{1+x^6}=\sum_\text{cyc}\sqrt{x^2(y^2z^2+x^4)}\le\sqrt{\Bigl(\sum_\text{cyc}x^2\Bigr)\Bigl(\sum_\text{cyc}(y^2z^2+x^4)\Bigr)}$$
So it suffices to prove
$$\Bigl(\sum_\text{cyc}x^2\Bigr)\Bigl(\sum_\text{cyc}(y^2z^2+x^4)\Bigr)\le 2\Bigl(\sum_\text{cyc}x^3\Bigr)^{\!2}$$
Whose expanded and eliminated version is
$$2\sum_\text{cyc}x^4y^2+2\sum_\text{cyc}x^2y^4+3x^2y^2z^2\le \sum_\text{cyc}x^6+4\sum_\text{cyc}x^3y^3$$
Which is after symmetrization equivalent to
$$0\le \Bigl(\sum_\text{sym}x^6-\sum_\text{sym}x^4y^2\Bigr)+3\Bigl(\sum_\text{sym}x^3y^3-\sum_\text{sym}x^4y^2\Bigr)+\Bigl(\sum_\text{sym}x^3y^3-\sum_\text{sym}x^2y^2z^2\Bigr)$$
The last difference is just AM-GM, so we can forget it. Now observe that
\begin{align*}\sum_\text{sym}x^6-\sum_\text{sym}x^4y^2&=\sum_\text{sym}(x-y)x^4(x+y)=\sum_\text{cyc}(x-y)(x^4-y^4)(x+y)\\&=\sum_\text{cyc}(x-y)^2(x^3+x^2y+xy^2+y^3)(x+y)\\
\sum_\text{sym}x^4y^2-\sum_\text{sym}x^3y^3&=\sum_\text{sym}(x-y)x^3y^2=\sum_\text{cyc}(x-y)(x^3y^2-x^2y^3)\\&=\sum_\text{cyc}(x-y)^2x^2y^2\end{align*}
So we are done if $0\le (x^3+x^2y+xy^2+y^3)(x+y)-3x^2y^2$ and one can easily check it's the same as
$$0\le (x^2-y^2)^2+x^2y^2+2x^3y+2xy^3$$
