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.

  • $\begingroup$ Where is this inequality from? $\endgroup$ May 28, 2012 at 15:10

1 Answer 1


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 $$\sum_{j=1}^n\frac{x_j}{(1+s_j)^2}\leq 1-\frac 1{1+s_n}<1.$$


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