Let $a_1,a_2,\ldots,a_n$ be real positive numbers. Prove that $$\left(1+\frac{a_1^2}{a_2}\right)\left(1+\frac{a_2^2}{a_3}\right) \cdots \left(1+\frac{a_n^2}{a_1}\right) \geq(1+a_1)(1+a_2) \cdots (1+a_n)$$

  • $\begingroup$ Where did this question come from? $\endgroup$ – Peter Woolfitt Jul 9 '16 at 18:39
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    $\begingroup$ There are solutions online, at least in my exotic language. Hint: prove first that $1+\frac{x^2}{y} \geq \frac{(1+x)^2}{1+y}$. $\endgroup$ – Emolga Jul 9 '16 at 18:41
  • $\begingroup$ 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 Deiga Jul 9 '16 at 18:43

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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    $\begingroup$ $$ABCDEFG$$ ... $$HIJKLMNOP$$ 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. $\qquad$ $\endgroup$ – Michael Hardy Jul 9 '16 at 19:42
  • $\begingroup$ @MichaelHardy Thanks for editing. $\endgroup$ – alans Jul 9 '16 at 19:47

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