Let $a_1,a_2,\ldots,a_n$ be real positive numbers. Prove that $$\left(1+\frac{a_1^2}{a_2}\right)\left(1+\frac{a_2^2}{a_3}\right) \cdots \left(1+\frac{a_n^2}{a_1}\right) \geq(1+a_1)(1+a_2) \cdots (1+a_n)$$
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$\begingroup$ Where did this question come from? $\endgroup$– Peter WoolfittJul 9, 2016 at 18:39
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3$\begingroup$ There are solutions online, at least in my exotic language. Hint: prove first that $1+\frac{x^2}{y} \geq \frac{(1+x)^2}{1+y}$. $\endgroup$– EmolgaJul 9, 2016 at 18:41
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$\begingroup$ It's from a list of problems of POTI (polo olímpico de treinamento intensivo). POTI is a brazillian program of olympic treinament of the IMPA. $\endgroup$– Rafael DeigaJul 9, 2016 at 18:43
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1 Answer
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By Cauchy-Schwarz inequality, we have the following: $$(1+a_2)\left(1+\frac{a_1^2}{a_2}\right)\geq (1+a_1)^2$$ $$(1+a_3)\left(1+\frac{a_2^2}{a_3}\right)\geq (1+a_2)^2$$ $$\vdots$$ $$(1+a_1)\left(1+\frac{a_n^2}{a_1}\right)\geq (1+a_n)^2$$ from where we have: $$\left(1+\frac{a_1^2}{a_2}\right)\left(1+\frac{a_2^2}{a_3}\right)\cdots\left(1+\frac{a_n^2}{a_1}\right)\prod_{i=1}^n (1+a_i)\geq \prod_{i=1}^n (1+a_i)^2.$$ By division with $\prod_{i=1}^n (1+a_i)$ we get $$\left(1+\frac{a_1^2}{a_2}\right)\left(1+\frac{a_2^2}{a_3}\right)\cdots\left(1+\frac{a_n^2}{a_1}\right)\geq \prod_{i=1}^n (1+a_i).$$
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2$\begingroup$ $$ABCDEFG$$ ... $$HIJKLMNOP$$ Contrast the above, with its three dots at the left edge of the page, with this: $$ABCDEFG$$ $$\vdots$$ $$HIJKLMNOP$$ The latter is the way to do it. (Except that in a more fastidious setting you'd probably use "align" or "gather" or something like that.) I edited the answer accordingly. $\qquad$ $\endgroup$ Jul 9, 2016 at 19:42
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