Inequality $\sqrt{1+x^2}+\sqrt{1+y^2}+\sqrt{1+z^2} \le \sqrt{2}(x+y+z)$

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.

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.

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 $$0\le (x^2-y^2)^2+x^2y^2+2x^3y+2xy^3$$

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!