Question:
Given $a,b,c,d > 0$, show that
$$
\frac{1}{4} \left(
\frac{a^2}{b} + \frac{b^2}{c} + \frac{c^2}{d} + \frac{d^2}{a}
\right) \ge \sqrt[4]{ \frac{a^4 + b^4 + c^4 + d^4 }{4} }
$$
Write
$$
Q(x_1,x_2,x_3,x_4) = \frac{1}{4} \left(
\frac{x_1^2}{x_2} + \frac{x_2^2}{x_3} + \frac{x_3^2}{x_4} + \frac{x_4^2}{x_1}
\right) - \left( \frac{ x_1^4 + x_2^4 + x_3^4 + x_4^4}{4} \right)^\tfrac{1}{4}
$$
Find extreme value for $Q$:
$$
\frac{\partial Q}{\partial x_k} = \frac{1}{2} \frac{x_k}{x_{|k+1|}}
- \frac{1}{4} \frac{x_{|k-1|}^2}{x_k^2}
- \frac{1}{4} x_k^3 \left( \frac{ x_1^4 + x_2^4 + x_3^4 + x_4^4}{4} \right)^{-\tfrac{3}{4}} = 0,\tag{1}
$$
where $x_{|k\pm1|}$ is the cyclic index, thus $x_{|0|} = x_4$ and $x_{|5|} = x_1$. The extreme value is obtained for $x_1=x_2=x_3=x_4=x$ and
$$
Q(x,x,x,x) = 0.
$$
We also have
$$
\frac{\partial^2 Q}{\partial x_k^2} = \frac{1}{2} \frac{1}{x_{|k+1|}}
+ \frac{1}{2} \frac{x_{|k-1|}^2}{x_k^3}
+ \frac{3}{16} x_k^6 \left( \frac{ x_1^4 + x_2^4 + x_3^4 + x_4^4}{4} \right)^{-\tfrac{7}{4}}
- \frac{3}{4} x_k^2 \left( \frac{ x_1^4 + x_2^4 + x_3^4 + x_4^4}{4} \right)^{-\tfrac{3}{4}},
$$
and for $x_1=x_2=x_3=x_4=x$ we obtain
$$
\frac{\partial^2 Q}{\partial x_k^2} = \frac{7}{16} \frac{1}{x} \ge 0,
$$
therefore
$$
Q(x_1,x_2,x_3,x_4) \ge 0,
$$
whence
$$\frac{1}{4} \left(
\frac{x_1^2}{x_2} + \frac{x_2^2}{x_3} + \frac{x_3^2}{x_4} + \frac{x_4^2}{x_1}
\right) \ge \left( \frac{ x_1^4 + x_2^4 + x_3^4 + x_4^4}{4} \right)^\tfrac{1}{4}.
$$
To show that the stationary solution is unique.
From equation (1) follows that
$$
2 \frac{x_{k+1}}{x_{k+2}} - \frac{x_{k}^2}{x_{k+1}^2} =
\left( 2 \frac{x_{k+2}}{x_{k+3}} - \frac{x_{k+1}^2}{x_{k+2}^2} \right)
\frac{x_{k+1}^3}{x_{k+2}^3}.
$$
Let us define
$$
\xi_k = \frac{x_k}{x_{k+1}},
$$
then we obtain
$$
2 \xi_{k+1} - \xi_k^2 = \Big( 2 \xi_{k+2} - \xi_{k+1}^2 \Big) \xi_{k+1}^3,
$$
and it is clear that $\xi_k = 1$ is a solution.
Let us write
$$
\xi_k = 1 + \phi_k,
$$
then we obtain
$$
\phi_{k+2} = \frac{ \phi_{k+1}^5 + 5 \phi_{k+1}^4 + 10 \phi_{k+1}^3
+ 8 \phi_{k+1}^2 + 3 \phi_{k+1}
- \phi_{k}^2 - 2 \phi_k}{ \Big( 1 + \phi_{k+1} \Big)^2 }.\tag{2}
$$
Note that for $\phi_\jmath > -1$, we obtain
$$
\begin{eqnarray}
\textrm{sgn}\Big( \phi_{k+1}^5 + 5 \phi_{k+1}^4 + 10 \phi_{k+1}^3
+ 8 \phi_{k+1}^2 + 3 \phi_{k+1} \Big) &=& \textrm{sgn}\Big(\phi_{k+1}\Big),\\
\textrm{sgn}\Big( \phi_{k}^2 - 2 \phi_k \Big) &=& \textrm{sgn}\Big(\phi_k\Big),
\end{eqnarray}
$$
whence we have the properties
$$
\begin{array}{ccccc}
\phi_k \le 0 &\wedge& \phi_{k+1} \ge 0 &\Rightarrow& \phi_{k+2} \ge 0,\\
\phi_k \ge 0 &\wedge& \phi_{k+1} \le 0 &\Rightarrow& \phi_{k+2} \le 0.\\
\end{array}
$$
We can use these properties to find "valid cycles":
$$
\begin{array}{ccccccccccc}
\phi_k = 0 &&
\phi_{k+1} = 0 &\Rightarrow&
\phi_{k+2} = 0 &\Rightarrow&
\phi_{k+3} = 0\\
\phi_k = 0 &&
\color{red}{\phi_{k+1} < 0} &\Rightarrow&
\phi_{k+2} < 0 &&
\phi_{k+3} < 0 &&
\phi_k = 0 &\Rightarrow&
\color{red}{\phi_{k+1} > 0}\\
\color{red}{\phi_k = 0} &&
\phi_{k+1} < 0 &\Rightarrow&
\phi_{k+2} < 0 &&
\phi_{k+3} > 0 &\Rightarrow&
\color{red}{\phi_k > 0}\\
\color{red}{\phi_k = 0} &&
\phi_{k+1} > 0 &\Rightarrow&
\phi_{k+2} > 0 &&
\phi_{k+3} < 0 &\Rightarrow&
\color{red}{\phi_k < 0}\\
\phi_k = 0 &&
\color{red}{\phi_{k+1} > 0} &\Rightarrow&
\phi_{k+2} > 0 &&
\phi_{k+3} > 0 &&
\phi_k = 0 &\Rightarrow&
\color{red}{\phi_{k+1} < 0}\\
\hline
\phi_k < 0 &&
\phi_{k+1} < 0 &&
\phi_{k+2} < 0 &&
\phi_{k+3} < 0\\
\color{red}{\phi_k < 0} &&
\phi_{k+1} < 0 &&
\phi_{k+2} < 0 &&
\phi_{k+3} > 0 &\Rightarrow&
\color{red}{\phi_k > 0}\\
\phi_k < 0 &&
\phi_{k+1} < 0 &&
\phi_{k+2} > 0 &\Rightarrow&
\phi_{k+3} > 0\\
\phi_k < 0 &&
\color{red}{\phi_{k+1} > 0} &\Rightarrow&
\phi_{k+2} > 0 &&
\phi_{k+3} > 0 &&
\phi_{k} < 0 &\Rightarrow&
\color{red}{\phi_{k+1} < 0}\\
\phi_k > 0 &&
\color{red}{\phi_{k+1} < 0} &\Rightarrow&
\phi_{k+2} < 0 &&
\phi_{k+3} < 0 &&
\phi_{k} > 0 &\Rightarrow&
\color{red}{\phi_{k+1} > 0}\\
\phi_k > 0 &&
\phi_{k+1} < 0 &\Rightarrow&
\phi_{k+2} < 0 &&
\phi_{k+3} > 0 &\Rightarrow&
\phi_k > 0\\
\phi_k > 0 &&
\phi_{k+1} > 0 &&
\phi_{k+2} < 0 &\Rightarrow&
\phi_{k+3} < 0\\
\color{red}{\phi_k > 0} &&
\phi_{k+1} > 0 &&
\phi_{k+2} > 0 &&
\phi_{k+3} < 0 &\Rightarrow&
\color{red}{\phi_k < 0}\\
\phi_k > 0 &&
\phi_{k+1} > 0 &&
\phi_{k+2} > 0 &&
\phi_{k+3} > 0\\
\end{array}
$$
Therefore we only have the following valid cycles
$$
\begin{array}{cccc}
0&0&0&0\\
-&-&-&-\\
-&-&+&+\\
+&+&+&+\\
\end{array}
$$
However the cycles
$$
\begin{array}{cccc}
-&-&-&-\\
+&+&+&+\\
\end{array}
$$
can be excluded for $\xi_1 \xi_2 \xi_3 \xi_4 = 1$.
What is left to show is that the cycle $--++$ leads to contradiction.
To be continued...
Best Answer
$$\left(1+\frac{x}{y}\right)\left(1+\frac{y}{z}\right)\left(1+\frac{z}{x}\right) =2+\left(\frac{x}{y}+\frac{y}{z}+\frac{z}{x}\right)+\left(\frac{y}{x}+\frac{z}{y}+\frac{x}{z}\right).$$
And each term in parentheses satisfies the inequality you have proved.