Prove $\sqrt[2]{x+y}+\sqrt[3]{y+z}+\sqrt[4]{z+x} <4$ 
$x,y,z \geqslant 0$ and $x^2+y^2+z^2+xyz=4$, prove
  $$\sqrt[2]{x+y}+\sqrt[3]{y+z}+\sqrt[4]{z+x} <4$$

A natural though is that from the condition $x^2+y^2+z^2+xyz=4$, I tried a trig substitutions $x=2\cdot \cos A$, $y=2\cdot \cos B$ and $z=2\cdot \cos C$, where $A,B,C$ are angles of an acute triangle. 
Our inequality becomes
$$\sqrt[2]{2(\cos A+ \cos B)}+\sqrt[3]{2(\cos B+\cos C)}+\sqrt[4]{2(\cos C+\cos A)} <4$$
I used formula $\cos(A)+\cos(B) < 2$ and estimate that $LHS <5$ but not $4$.
I do not know how to proceed. 
 A: The condition $\;x^2+y^2+z^2+xyz=4\;$ can be rewritten as:
$$
z^2+xy\cdot z + (x^2+y^2-4) = 0 \quad \Longrightarrow \quad
z = - \frac{xy}{2} \pm \sqrt{\left(\frac{xy}{2}\right)^2-(x^2+y^2-4)}
$$
From $z \ge 0$ it follows that:
$$
z = - \frac{xy}{2} + \sqrt{\left(\frac{xy}{2}\right)^2-(x^2+y^2-4)} \quad \mbox{and} \quad x^2+y^2\le 4
$$
Therefore, after substitution of $\,z$ ,
we only have to investigate function behavior$f(x,y) = \sqrt[2]{x+y}+\sqrt[3]{y+z}+\sqrt[4]{z+x} < 4$
in the area enclosed by x-axis, y-axis and a quarter of a circle, as depicted below.

Another proof without words is attempted
by plotting a contour map of the function, as depicted. Levels (nivo) of these isolines are defined
(in Delphi Pascal) as:

nivo := min + g/grens*(max-min); { grens = 40 ; g = 1..grens }

The blackness of the isolines is proportional to the (positive) function values; they are almost white
near the minimum and almost black near the maximum values.
Maximum and minimum values of the function are observed to be:


 2.54387743763872E+0000 < f < 3.91477606446737E+0000

The $\color{blue}{\mbox{blue}}$ spot is where $\,\left| f(x,y) - 4 \right| < 0.9$ .
It is suggested by the rather large error $\,0.9\,$ that $\,4\,$ is not really the maximum.
Indeed, upon refinement of the grid we find for the maximum numerically (double precision)
a somewhat lower value:


3.91477205860402 < 4

A: Remark: My second proof is given in https://artofproblemsolving.com/community/c6h2483725p20868007.
There, KaiRain@AoPS gave a nice proof.
Proof without using derivative
We split into three cases:

*

*$y+z \le 1$ and $z+x \le 1$: We have $x + y \le 2$.
Thus, we have $\mathrm{LHS} \le \sqrt{2} + 1 + 1 < 4$. The inequality is true.


*$y + z > 1$ and $z + x \le 1$: Since $\sqrt[3]{y+z} < \sqrt{y+z}$ and $\sqrt[4]{z+x} \le 1$, it suffices to prove that
$$\sqrt{x+y} + \sqrt{y+z} \le 3.$$
By AM-QM inequality, we have $\sqrt{x+y} + \sqrt{y+z} \le \sqrt{2(x+y + y + z)}$. Thus,
It suffices to prove that $$x+2y+z \le \frac{9}{2}.$$
From $x^2+y^2+z^2 + xyz = 4$, we have
$y^2 + (\frac{1}{2} + \frac{y}{4})(x + z)^2 + (\frac{1}{2} - \frac{y}{4})(x - z)^2 = 4$
which results in $y^2 + (\frac{1}{2} + \frac{y}{4})(x + z)^2 \le 4$
and $x + z \le \sqrt{\frac{4(4 - y^2)}{2 + y}}$.
Thus, it suffices to prove that $2y + \sqrt{\frac{4(4 - y^2)}{2 + y}} \le \frac{9}{2}$.
It suffices to prove that $(\frac{9}{2} - 2y)^2 \ge \frac{4(4 - y^2)}{2 + y}$
that is $\frac{1}{4}(4y - 7)^2 \ge 0$. The inequality is true.


*$z + x > 1$: Since $\sqrt[3]{z+x} > \sqrt[4]{z+x}$, it suffices to prove that
$$
\sqrt{x+y}  + \sqrt[3]{y+z} + \sqrt[3]{z+x} < 4.
$$
Note that $u\mapsto \sqrt[3]{u}$ is concave on $u > 0$.
It suffices to prove that
$$
\sqrt{x+y}  + 2\sqrt[3]{\frac{y+z + z+x}{2}} < 4.
$$
From $x^2+y^2+z^2 + xyz = 4$, we have $(z + \frac{1}{2}xy)^2 = \frac{1}{4}(4-x^2)(4-y^2)$
which results in $$z = \frac{1}{2}\sqrt{(4-x^2)(4-y^2)}-\frac{1}{2}xy \le \frac{1}{2}\frac{4-x^2+4-y^2}{2} - \frac{1}{2}xy  = 2- \frac{(x+y)^2}{4}.$$
Thus, it suffices to prove that
$$
\sqrt{x+y}  + 2\sqrt[3]{\frac{y+x}{2} + 2- \frac{(x+y)^2}{4}} < 4.
$$
Let $v = \sqrt{x+y}$. Since $x+y \le \sqrt{2(x^2+y^2)} \le \sqrt{2 \cdot 4} < 4$, we have $v\in (0, 2)$.
It suffices to prove that $$v + 2\sqrt[3]{\frac{v^2}{2} + 2- \frac{v^4}{4}} < 4$$
or $$\frac{v^2}{2} + 2- \frac{v^4}{4} < \frac{1}{8}(4 - v)^3$$ or $$\frac{1}{8}(2v^4-v^3+8v^2-48v+48) > 0$$ for $v\in (0, 2)$.
It is easy to prove that $2v^4-v^3+8v^2-48v+48 > 0$ for $v\in (0, 2)$.
We are done.
A: PARTIAL SOLUTION 
$$f(x,y,z)=\sqrt[2]{x+y}+\sqrt[3]{y+z}+\sqrt[4]{z+x}$$
$$g(x,y,z)= x^2+y^2+z^2+xyz-4=0$$
$$ \begin{align}\mathcal{L}(x,y,z,\lambda) =& \;f(x,y,z) + \lambda g(x,y,z)\\
=&\;\sqrt[2]{x+y}+\sqrt[3]{y+z}+\sqrt[4]{z+x}+\lambda(x^2+y^2+z^2+xyz-4) \end{align}$$
$$\nabla_{x,y,z,\lambda}\mathcal{L}(x,y,z,\lambda) =\left(\frac{\partial \mathcal{L}}{\partial x},\frac{\partial \mathcal{L}}{\partial y},\frac{\partial \mathcal{L}}{\partial z},\frac{\partial \mathcal{L}}{\partial \lambda}\right)$$
Now comes the work finding the partial derivatives
$$\begin{align}
&\frac{\partial \mathcal{L}}{\partial x} =\; \lambda(2x+yz) + \frac{1}{2(x+y)^{1/2}}+\frac{1}{4(x+z)^{3/4}}\\
&\frac{\partial \mathcal{L}}{\partial y} =\; \lambda(2y+xz) + \frac{1}{2(x+y)^{1/2}}+\frac{1}{3(y+z)^{2/3}}\\
&\frac{\partial \mathcal{L}}{\partial z} =\; \lambda(2z+xy) + \frac{1}{3(y+z)^{2/3}}+\frac{1}{4(x+z)^{3/4}}\\
&\frac{\partial \mathcal{L}}{\partial \lambda} =\; x^2+y^2+z^2+xyz-4
\end{align}$$  
$$\nabla_{x,y,z,\lambda}\mathcal{L}(x,y,z,\lambda)=0 \iff \begin{cases}
\lambda(2x+yz) + \frac{1}{2(x+y)^{1/2}}+\frac{1}{4(x+z)^{3/4}}  & = 0 \\
\lambda(2y+xz) + \frac{1}{2(x+y)^{1/2}}+\frac{1}{3(y+z)^{2/3}} & = 0 \\
\lambda(2z+xy) + \frac{1}{3(y+z)^{2/3}}+\frac{1}{4(x+z)^{3/4}}  & = 0 \\
x^2+y^2+z^2+xyz-4, & = 0
\end{cases}$$
At this point I cannot seem to solve the resulting equations. The symmetry involved implies some simple trick should suffice, though I have yet to spot one.
