7
$\begingroup$

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

$\endgroup$
3
  • $\begingroup$ Where did this question come from? $\endgroup$ – Peter Woolfitt Jul 9 '16 at 18:39
  • 3
    $\begingroup$ There are solutions online, at least in my exotic language. Hint: prove first that $1+\frac{x^2}{y} \geq \frac{(1+x)^2}{1+y}$. $\endgroup$ – 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
12
$\begingroup$

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 $$\left(1+\frac{a_1^2}{a_2}\right)\left(1+\frac{a_2^2}{a_3}\right)\cdots\left(1+\frac{a_n^2}{a_1}\right)\geq \prod_{i=1}^n (1+a_i).$$

$\endgroup$
2
  • 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. $\qquad$ $\endgroup$ – Michael Hardy Jul 9 '16 at 19:42
  • $\begingroup$ @MichaelHardy Thanks for editing. $\endgroup$ – alans Jul 9 '16 at 19:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.