<s>[INST] 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. 
 [/INST] 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.
</s>