If $x_1,x_2,\ldots,x_n$ are real numbers larger than $1$, prove that $$\dfrac{1}{1+x_1}+\dfrac{1}{1+x_2}+\cdots+\dfrac{1}{1+x_n} \geq \dfrac{n}{\sqrt[n]{x_1x_2\cdots x_n}+1}$$


AM-GM doesn't work here since we will get an upper bound. I don't see Cauchy-Schwarz working either. Thus, I think a substitution might work, but I am unsure of which one to use.

  • $\begingroup$ Are you sure of this inequality ? For $n=1$, $x_1=2$ it's wrong ! But it is true for $n=2$ and $x_1=2$, $x_2=3$... $\endgroup$
    – Tom-Tom
    Jan 4, 2016 at 20:23
  • $\begingroup$ So sorry, a typo. $\endgroup$ Jan 4, 2016 at 20:25
  • $\begingroup$ Much better this way ! What about the concavity of $x\mapsto \frac1{1+x}$ combined with AM-GM inequality ? $\endgroup$
    – Tom-Tom
    Jan 4, 2016 at 20:26
  • $\begingroup$ How would you use concavity? $\endgroup$ Jan 4, 2016 at 20:30
  • $\begingroup$ The answer below reflects what I had in mind. $\endgroup$
    – Tom-Tom
    Jan 4, 2016 at 20:53

2 Answers 2


Consider the function $f(x)=\frac{1}{1+e^x}$ which is convex for $x>0$ .

Now use Jensen's inequality :

$$f( \ln x_1)+f( \ln x_2)+\ldots+f( \ln x_n) \geq n f \left (\frac{\ln x_1+\ln x_2+\ldots+\ln x_n}{n} \right)$$

This is exactly your inequality :

$$\frac{1}{1+x_1}+\frac{1}{1+x_2}+\ldots+\frac{1}{1+x_n} \geq \frac{n}{1+\sqrt[n]{x_1x_2\ldots x_n}}$$


Use Cauchy : $$\frac{1}{1+x_{1}} + ... >= \frac{n}{((1+x_{1})...)^{\frac{1}{n}}}$$

Consider the: $$(1+x_{1})...)^{\frac{1}{n}} <= (x_{1}...)^{\frac{1}{n}}+1$$

The last one you could prove by yourself(use induction).

  • $\begingroup$ $\sqrt{(1+a)(1+b)} \ge \sqrt{ab}+1$, that is why OP said that Cauchy doesn't work. $\endgroup$
    – chenbai
    Jan 5, 2016 at 0:42
  • $\begingroup$ @chenbai but I used Cauchy $\endgroup$
    – openspace
    Jan 5, 2016 at 9:14
  • $\begingroup$ yes, but the result you got is weak, and your last step is in wrong direction which I show you in a simple example. $\endgroup$
    – chenbai
    Jan 5, 2016 at 12:32

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