<s>[INST] 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 ?
 [/INST] Consider the function $f(t) = \sqrt2t - \sqrt{t^2+1}-\left(\sqrt2-\frac1{\sqrt2} \right)\log t$ for $t > 0$. Note that $f(t) \ge 0 \implies f(x)+f(y)+f(z) \ge 0 $ which gives the desired inequality.  
$$$$
Now $f'(t) = \dfrac{-2t^2+2t\sqrt{2(t^2+1)}-\sqrt{2(t^2+1)}}{2t\sqrt{t^2+1}}$ so $f'(1)=0$.  Further,  
$$f''(t) = \frac{(t^2+1)^{3/2}- \sqrt2 t^2}{\sqrt 2 t^2 (t^2+1)^{3/2}} > 0$$
so $f'(t)> 0$ for $t> 1$ and $f'(t) < 0$ for $t < 1$. Hence $f(1)=0$ is the minimum, $f(t) \ge 0 \quad \forall t > 0$ and the inequality holds.

P.S.: @user1537366 when the inequality is separable, i.e. of form $g(x)+g(y)+g(z)+... \ge 0$ and a constraint which can be also expressed as $h(x)+h(y)+\dots = 0$, it is worthwhile to check if for a suitable constant $k$, we can get $f(t) = g(t)+k\,h(t) \ge 0$.  
Often $k$ can be determined by setting $f'(t^*)=0$ where $t=t^*$ is when you expect equality.
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