Help w/$(\frac{a}{b})^4+(\frac{b}{c})^4+(\frac{c}{d})^4+(\frac{d}{e})^4+(\frac{e}{a})^4\ge\frac{b}{a}+\frac{c}{b}+\frac{d}{c}+\frac{e}{d}+\frac{a}{e}$ How exactly do I solve this problem? (Source: 1984 British Math Olympiad #3 part II)
\begin{equation*}
\bigl(\frac{a}{b}\bigr)^4 + \bigl(\frac{b}{c}\bigr)^4 + \bigl(\frac{c}{d}\bigr)^4 + \bigl(\frac{d}{e}\bigr)^4 + \bigl(\frac{e}{a}\bigr)^4 \ge \frac{b}{a} + \frac{c}{b} + \frac{d}{c} + \frac{e}{d} + \frac{a}{e}
\end{equation*}
There's not really a clear-cut way to use AM-GM on this problem. I've been thinking of maybe using the Power Mean Inequality, but I don't exactly see a way to do that. Maybe we could use harmonic mean for the RHS?
 A: Applying the AM-GM
$$LHS - \bigl(\frac{e}{a}\bigr)^4 = \bigl(\frac{a}{b}\bigr)^4 + \bigl(\frac{b}{c}\bigr)^4 + \bigl(\frac{c}{d}\bigr)^4 + \bigl(\frac{d}{e}\bigr)^4  \ge4 \cdot \frac{a}{b}  \cdot \frac{b}{c}   \cdot \frac{c}{d}  \cdot\frac{d}{e} = 4\cdot\frac{a}{e} $$
Do the same thing for these 4 others terms, and make the sum
$$5 LHS - LHS \ge  4 RHS$$
$$\Longleftrightarrow LHS \ge RHS$$
The equality occurs when $a=b=c=d=e$
A: When you know there are some downvoters, you try some  other (Lagrangian) methods. (Not a complete solution. Hard part remains.)
What is the minumum of the function $f(x_1,x_2,x_3,x_4,x_5)=\sum_{i=1}^{n}x_i^4-x_i^{-1}$ with domain $\Bbb{R}^{5+}$, subject to the constraint equation $x_1x_2x_3x_4x_5=1$.
The system of a Lagrange multplier $\lambda$ gives the equations $4x_i^3+x_i^{-2}=\lambda$ for all $i=1,2,3,4,5$. From these equations and the constraint we deduce that $\lambda\geq 5$.
We observe with the help of WolframAlpha that a choice of the solution of the system of equations $4x_i^5-\lambda x_i^2+1=0$ can satisfy the constraint when $\lambda=5$ and all $x_i=1$.
See for example: https://www.wolframalpha.com/input?i=y%3D4x%5E5-5.01x%5E2%2B1
