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The question is as follows:

Let $a_1 \geq a_2 \geq \ldots \geq a_n > 0$ be n real numbers. Prove that $$ {a_1 \over a_2} + {a_2 \over a_3} + \ldots + {a_{n-1} \over a_n} + {a_n \over a_1} \leq {a_2 \over a_1} + {a_3 \over a_2} + \ldots + {a_n \over a_{n-1}} + {a_1 \over a_n}$$

I've tried grouping the terms and rearranging them. But, nothing so far. It'd be great if someone could help.

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I give prove for $n=3$. The proof in the general case is similar. So, we prove that $$ \frac{a_1}{a_2}-\frac{a_2}{a_1} + \frac{a_2}{a_3}-\frac{a_3}{a_2}\leq \frac{a_1}{a_3}-\frac{a_3}{a_1} $$ or $$ (a_1-a_2)\left(\frac{1}{a_1} + \frac{1}{a_2}\right)/2+ (a_2-a_3)\left(\frac{1}{a_2} + \frac{1}{a_3}\right)/2\leq (a_1-a_3)\left(\frac{1}{a_1} + \frac{1}{a_3}\right)/2. $$ Idea: each summand is the square of trapezium, e.g., $(a_1-a_2)\left(\frac{1}{a_1} + \frac{1}{a_2}\right)/2$ is the square of a trapezium with the height $(a_1-a_2)$ and the parallel sides $\frac{1}{a_1}$ and $\frac{1}{a_2}$. Now, the last inequality follows from the fact that the function $\frac{1}{x}$ is convex. (The trapezium in the RHS contains two trapeziums in the LHS)

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  • $\begingroup$ Thanks a lot. I was able to build up the proof using it. $\endgroup$ – SinTan1729 Aug 28 '16 at 10:14

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