If $a, b, c, d \in [\frac12, 2]$ and $abcd =1$, find the maximum value of $$(a+\frac1b)(b+\frac1c)(c+\frac1d)(d+\frac1a)$$

Thought: $$\begin{split}(a+\frac1b)(b+\frac1c)(c+\frac1d)(d+\frac1a) &= \frac{(ab+1)(bc+1)(cd+1)(da+1)}{abcd} \\ & \stackrel{\text{C-S}}{\le} \sqrt{(a^2+1)(b^2+1)}\sqrt{(b^2+1)(c^2+1)}\sqrt{(c^2+1)(d^2+1)}\sqrt{(d^2+1)(a^2+1)} \\ & = (a^2+1)(b^2+1)(c^2+1)(d^2+1)\end{split}$$

  • $\begingroup$ I got my AM-GM inequality confused. Deleted now. $\endgroup$
    – abnry
    Feb 17, 2016 at 20:01

2 Answers 2


This problem is not new: I have found it together with a solution (written in latex + chinese) in :


Here is one of the interesting solutions presented there, with a slight personal adaptation.

We begin by a lemma that will be used in the sequel:

The convex function defined by $f(x)=x+\frac{1}{x}$ on interval $[\frac{1}{2},2]$ reaches its maximum values at the limit points of its domain, i.e. for $x=1/2$ or $x= 2$, with $f(1/2)=f(2)=2.5$.

Now for the proof. Let us remind the constraint $abcd=1 \ \ \ (1)$

We start from $E=\frac{(ab+1)(bc+1)(cd+1)(da+1)}{abcd} $.

Taking into account (1) and changing the previous factor order into:


We expand inside each bracketed expression ; using (1), we obtain:

$E=(\sqrt{ab}+\sqrt{cd})^2 (\sqrt{bc}+\sqrt{ad})^2 \ \ \ (2)$

Let $m=\sqrt{ab}$ and $n=\sqrt{bc}$. Using (1), we have:

$E=\left(\left(m+\frac{1}{m}\right)\left(n+\frac{1}{n}\right)\right)^2=\left(mn+\frac{1}{mn}+\frac{m}{n}+\frac{n}{m}\right)^2 = (f(mn)+f(\frac{m}{n}))^2$

where $f$ is a notation introduced in the lemma ; we are in position to apply it (see Appendix explaining why $mn$ and $\frac{m}{n}$ are in $[0.5,2]$) ; it gives the final majorization:

$E \leq (2.5 + 2.5)^2=25 \ \ \ (5)$

Besides, taking

$a=2, b=2, c=\dfrac{1}{2}, d=\dfrac{1}{2}$

in expression $E$ yields this value $25$, which is thus the maximal value of expression $E$.

Remark : this solution uses neither the AGM, nor Cauchy-Schwarz inequality.

APPENDIX: The domains of $\frac{m}{n}$ and $mn$.

Using (1), $\frac{m}{n}=\sqrt{\frac{a}{c}}$. But $\frac{1}{2} \leq a \leq 2$ and $\frac{1}{2} \leq c \leq 2$. Thus $\frac{m}{n} \in [\frac{1}{2},2]$.

$mn=\sqrt{abc}\sqrt{b}=\sqrt{\frac{1}{d}}\sqrt{b}=\sqrt{\frac{b}{d}}$. Thus by a similar reasoning as before, $mn \in [\frac{1}{2}, 2]$.


I think it can help:


See an elim's proof.


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