8
$\begingroup$

Let be $x_1, x_2, \ldots , x_n$ strictly positive real numbers. Prove that the following inequality holds:

$$\frac1{1+x_1}+\frac1{1+x_1+x_2}+\cdots+\frac1{1+x_1+x_2+\cdots+x_n} < \sqrt{\frac1{x_{1}}+\frac1{x_2}+\cdots+\frac1{x_n}}$$

How may i tackle this inequality? I tried AM-GM, but it seems of no help.

$\endgroup$
  • $\begingroup$ Where is this inequality from? $\endgroup$ – Davide Giraudo May 28 '12 at 15:10
10
$\begingroup$

Let $s_j:=\sum_{k=1}^jx_j$. We have by Cauchy-Schwarz inequality that $$\sum_{j=1}^n\frac 1{1+s_j}\leq \sqrt{\sum_{j=1}^n\frac{x_j}{(1+s_j)^2}}\cdot \sqrt{\sum_{j=1}^n\frac 1{x_j}},$$ so it remains to show that $\sum_{j=1}^n\frac{x_j}{(1+s_j)^2}<1$. We have $$\frac{x_j}{(1+s_j)^2}\leq \frac{x_j}{(1+s_j)(1+s_{j-1})}=\frac{(1+s_j)-(1+s_{j-1})}{(1+s_j)(1+s_{j-1})}=\frac 1{1+s_{j-1}}-\frac 1{1+s_j}$$ hence $$\sum_{j=1}^n\frac{x_j}{(1+s_j)^2}\leq 1-\frac 1{1+s_n}<1.$$

$\endgroup$

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.