# A nice inequality with positive real tuples

I have found out this very nice inequality, which I have not been able to solve and I think it is an interesting challenge. Simulation results seem to verify it.

Consider $n$ positive (strictly larger than zero) real numbers $x_1,x_2,...,x_n$ such that $x_1+x_2+...+x_n=1$ and $n$ other arbitrary positive (strictly larger than zero) real numbers, denoted as $y_1,y_2,...y_n$. Show that: $$\left(\frac{1}{\sum_{i=1}^n{\frac{x_i}{y_i}}}+1\right) \cdot \sum_{i=1}^n{\frac{x_i}{y_i+1}} \leq 1.$$

Let me know if you can help me.

• Have you tried proving it for $n = 1, 2, \ldots$? – Rob Arthan Aug 9 '17 at 21:20
• Yes, this was the first thing I tried. However, induction didn't look to work - however, maybe it is possible to make it working. Thanks a lot anyway ! – Enrico Piovano Aug 11 '17 at 17:24

We need to prove that $$\left(\frac{1}{\sum\limits_{i=1}^n{\frac{x_i}{y_i}}}+1\right) \sum_{i=1}^n{\frac{x_i}{y_i+1}} \leq 1$$ or $$\sum_{i=1}^n{\frac{x_i}{y_i+1}}\leq\frac{\sum\limits_{i=1}^n\frac{x_i}{y_1}}{1+\sum\limits_{i=1}^n\frac{x_i}{y_1}}$$ or $$\sum_{i=1}^n\left(\frac{x_i}{y_i+1}-x_i\right)\leq\frac{\sum\limits_{i=1}^n\frac{x_i}{y_1}}{1+\sum\limits_{i=1}^n\frac{x_i}{y_1}}-1$$ or $$\sum\limits_{i=1}^n\frac{x_iy_i}{y_i+1}\geq\frac{1}{1+\sum\limits_{i=1}^n\frac{x_i}{y_1}}$$ or $$\left(1+\sum\limits_{i=1}^n\frac{x_i}{y_1}\right)\sum\limits_{i=1}^n\frac{x_iy_i}{y_i+1}\geq1$$ or

$$\sum\limits_{i=1}^n\left(\frac{x_i}{y_1}+x_i\right)\sum\limits_{i=1}^n\frac{x_iy_i}{y_i+1}\geq1$$ or

$$\sum\limits_{i=1}^n\frac{x_i(y_i+1)}{y_1}\sum\limits_{i=1}^n\frac{x_iy_i}{y_i+1}\geq1,$$ which is C-S:

$$\sum\limits_{i=1}^n\frac{x_i(y_i+1)}{y_1}\sum\limits_{i=1}^n\frac{x_iy_i}{y_i+1}\geq\left(\sum\limits_{i=1}^nx_i\right)^2=1.$$ Done!

• Thanks a lot Michael! Your solution with Cauchy-Schwarz is very nice ! – Enrico Piovano Aug 11 '17 at 22:41

The result follows directly from Jensen's inequality applied to the function $u\mapsto u/(1+u),$ which is concave on $\mathbb R_+$. Define a random variable $U$ for which $U=1/y_i$ with probability $x_i$, so for instance the expectation of $U$ is $EU=\sum_{i=1}^n x_i/y_i$ and $E(U/(1+U))= \sum_{i=1}^n x_i /(1+y_i)$ and so on. Then according to Jensen, $$E\frac U {1+U}\le \frac{EU}{1+EU},$$ which is equivalent to $$\left( \frac 1{EU}+1\right ) \frac{EU}{1+EU}\le 1,$$ which is the OP's original inequality.

• @ Is $E$ the expected value? Nice solution! +1. We can write this solution without $E$, but with $E$ it's much nicer! – Michael Rozenberg Aug 9 '17 at 22:09
• @MichaelRozenberg Yes. I have edited my answer to make this clearer. Thanks for the plug! – kimchi lover Aug 9 '17 at 22:17
• Thanks a lot! This application with Jensen's inequality is very brilliant! – Enrico Piovano Aug 11 '17 at 22:44