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


Attempt

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

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 :
$$\frac{1}{1+x_{1}} + ... &gt;= \frac{n}{((1+x_{1})...)^{\frac{1}{n}}}$$

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

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