$(\frac{a}{b})^4+(\frac{b}{c})^4+(\frac{c}{d})^4+(\frac{d}{e})^4+(\frac{e}{a})^4\ge\frac{b}{a}+\frac{c}{b}+\frac{d}{c}+\frac{e}{d}+\frac{a}{e}$

Question

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?

Answer 1

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 $a=b=c=d=e$

Answer 2

NN2 gave a simple and very elegamt proof. I tried another way.

What is the minumum of the function $f(x_1,x_2,x_3,x_4,x_5)=\sum_{i=1}^{5}(x_i^4-x_i^{-1})$ 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 $4x_i^3+x_i^{-2}=\lambda x_i^{-1}$ for all $i=1,2,3,4,5$. From these equations we have $$4x_ix_j(x_i^4-x_j^4)=x_i-x_j$$ for all $i,j.$ I am stuck. Any ideas?