<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] 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.
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