# Help w/$(\frac{a}{b})^4+(\frac{b}{c})^4+(\frac{c}{d})^4+(\frac{d}{e})^4+(\frac{e}{a})^4\ge\frac{b}{a}+\frac{c}{b}+\frac{d}{c}+\frac{e}{d}+\frac{a}{e}$ [closed]

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?

• someone please explain why this is closed. I think I have adequately explained some strategies that I've tried. I believe I've provided enough context. Dec 10, 2022 at 19:54
• I'm kinda new around here, but I was also surprised to see it closed. Also I found the accepted solution to be very nice. Dec 10, 2022 at 21:48

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 $$a=b=c=d=e$$

• I think in the end you should have $5LHS-LHS\geq 4RHS$, since you repeat the procedure 5 times, not 4. Then everything works. :) Dec 5, 2022 at 9:33
• @Freshman'sDream You're right, I just corrected this typo. Thanks!
– NN2
Dec 5, 2022 at 9:34
• ohhhhh ok thanks! Dec 9, 2022 at 23:08

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