# $\frac{2}{x_1+1} + \frac{2}{x_2+1} + \frac{2}{x_3+1} + ... + \frac{2}{x_n+1} \leq n$

Let: $$x_1,x_2,x_3,...,x_n>0$$ and $$x_1x_2x_3...x_n=1$$. Prove that: $$x_1+x_2+x_3+...+x_n \geq \frac{2}{x_1+1} + \frac{2}{x_2+1} + \frac{2}{x_3+1} + ... + \frac{2}{x_n+1}$$

I used Cauchy in LHS:

$$LHS \geq n$$ so I have to prove:

$$\frac{2}{x_1+1} + \frac{2}{x_2+1} + \frac{2}{x_3+1} + ... + \frac{2}{x_n+1} \leq n$$

and it can be rewritten as:

$$\sum_{i=1}^{n} \frac{x_i-1}{1+x_i} \geq 0$$

I want to use Chebyshev but it was wrong!

• What else have you tried? Can you prove the case $n=2$? Commented Sep 29, 2023 at 15:25
• Have you tried induction? Commented Sep 29, 2023 at 15:32
• I think the case for $n=2$ is easy, but induction is not easy with this! Commented Sep 29, 2023 at 15:39
• The function $f\left(y\right)=e^{y}-\frac{2}{1+e^{y}}$ is convex with $f\left(0\right)=0$. Now apply Jensen’s inequality with $y_{i}=\log\left(x_{i}\right)$.
– Rafi
Commented Sep 29, 2023 at 18:01
• Can you use AM-GM? Commented Sep 29, 2023 at 23:04

Well $$\sum \dfrac2{x_i+1}\leqslant n$$ does not always hold true (take $$x_1=t^{n-1}, x_{i\neq1}=1/t$$ and let $$t\to \infty$$), so we work with the original problem.
It is enough to show $$f(x) = x-\frac2{x+1}-\frac32\log(x)\geqslant 0$$, as the original inequality is the same as $$\sum f(x_i) \geqslant 0$$.
However $$f'(x)=(x-1)\dfrac{2x^2+3x+3}{2x(x+1)^2}$$, hence $$f$$ is decreasing in $$(0,1)$$ and increasing for $$x>1$$. As $$f(1)=0$$, we must have $$f(x)\geqslant 0$$ for all positives, and so we are done.
• Congratulations for the clever choice of $f$ Commented Sep 29, 2023 at 18:42
$$\frac2{x+1}=2-\frac{2x}{x+1}$$ Hence the inequality can be equivalently written as $$\sum_i \left(x_i+\frac{2x_i}{x_i+1}\right) \geqslant 2n$$ $$\iff \sum_i \left(x_i+1+\frac{2x_i}{x_i+1}\right) \geqslant 3n$$ By AM-GM, we have $$\frac{x+1}2+\frac{2x}{x+1} \geqslant 2\sqrt{x}$$ Hence it is enough to show $$\sum_i \frac{x_i+1}2+2\sum_i \sqrt{x_i}\geqslant 3n$$ which follows from the AM-GMs $$\sum x_i \geqslant n\sqrt[n]{\prod x_i}=n$$ and $$\sum \sqrt{x_i} \geqslant n\sqrt[2n]{\prod x_i}=n$$.