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 ?

Let $x=\frac{a^2}{bc}$, $y=\frac{b^2}{ac}$ and $z=\frac{c^2}{ab}$, where $a$, $b$ and $c$ are positive numbers.
Hence, we need to prove that:
$$\sum_{cyc}a\sqrt{a^4+b^2c^2}\leq\sqrt2(a^3+b^3+c^3).$$
Now, by C-S
$$\left(\sum_{cyc}a\sqrt{a^4+b^2c^2}\right)^2\leq(a^2+b^2+c^2)\sum_{cyc}(a^4+b^2c^2)$$ and it's enough to prove that
$$2(a^3+b^3+c^3)^2\geq(a^2+b^2+c^2)\sum_{cyc}(a^4+b^2c^2)$$ or
$$\sum_{cyc}(a^6-2a^4b^2-2a^4c^2+4a^3b^3-a^2b^2c^2)\geq0$$ or
$$\sum_{cyc}(a-b)^2\left(\frac{(a^2+b^2+c^2)(a+b)^2}{2}-2a^2b^2\right)\geq0,$$
which is true because by AM-GM
$$\frac{(a^2+b^2+c^2)(a+b)^2}{2}\geq\frac{(a^2+b^2)(a+b)^2}{2}\geq ab\left(2\sqrt{ab}\right)^2=4a^2b^2>2a^2b^2.$$
Done!
