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 '16 at 20:23
  • $\begingroup$ So sorry, a typo. $\endgroup$ – Jacob Willis Jan 4 '16 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 '16 at 20:26
  • $\begingroup$ How would you use concavity? $\endgroup$ – Jacob Willis Jan 4 '16 at 20:30
  • $\begingroup$ The answer below reflects what I had in mind. $\endgroup$ – Tom-Tom Jan 4 '16 at 20:53

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 '16 at 0:42
  • $\begingroup$ @chenbai but I used Cauchy $\endgroup$ – openspace Jan 5 '16 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 '16 at 12:32

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.