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?

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    $\begingroup$ 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. $\endgroup$ Dec 10, 2022 at 19:54
  • $\begingroup$ 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. $\endgroup$
    – 3rdMoment
    Dec 10, 2022 at 21:48

2 Answers 2


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 $a=b=c=d=e$

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


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