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. 

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

