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How to prove the following inequality:

For $a_i\in(0,1)$ ($i=1,\cdots,N$), N>1, then $$\prod\limits_{i=1}^Na_i+\prod\limits_{i=1}^N\sqrt{1-a_i^2}\leq1.$$

Thank you, GUYS!

Maybe try considering some trigonometric functions?

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  • $\begingroup$ I think that for $N=1$ doesn't hold true. Take for example $a_1=\frac{1}{\sqrt{2}}$ $\endgroup$
    – foo90
    Nov 19, 2015 at 17:07
  • $\begingroup$ Yes, I add a patch N>1. Thank you. $\endgroup$
    – peter
    Nov 19, 2015 at 17:15

1 Answer 1

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When $N=2$, this is Cauchy-Schwarz: if $a_1^2+b_1^2=1$ and $a_2^2+b_2^2=1$, then $$ a_1a_2+b_1b_2\leq (a_1^2+a_2^2)^{1/2}(b_1^2+b_2^2)^{1/2}=1. $$ This suggests a proof by induction: \begin{align} \prod_{j=1}^{N+1}a_j+\prod_{j=1}^{N+1}\sqrt{1-a_j^2}&=(a_{N+1},\sqrt{1-a_{N+1}^2})\cdot\left(\prod_{j=1}^{N}a_j,\prod_{j=1}^{N}\sqrt{1-a_j^2}\right)\\ \ \\ &\leq (a_{N+1}^2+{1-a_{N+1}^2})^{1/2}\,\left(\prod_{j=1}^{N}a_j^2+\prod_{j=1}^{N}(1-a_j^2)\right)^{1/2}\\ \ \\ &=\left(\prod_{j=1}^{N}a_j^2+\prod_{j=1}^{N}(1-a_j^2)\right)^{1/2}\\ \ \\ &\leq\left(\prod_{j=1}^{N}a_j+\prod_{j=1}^{N}\sqrt{1-a_j^2}\right)^{1/2}\\ \ \\ &\leq1, \end{align} where in the second to last inequality we use the fact that if $t\in (0,1]$ then $t^2\leq t$, and in the last inequality we use the inductive hypothesis.

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  • $\begingroup$ Thanks a million!! PS: In the second inequality, there is a small typo. But anyway, thank you very very much for your help. $\endgroup$
    – peter
    Nov 19, 2015 at 17:13
  • $\begingroup$ Good thing you noticed, and I wouldn't call it small! Corrected now. $\endgroup$ Nov 19, 2015 at 18:47

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