Show the following inequality holds I want to show that $( 1 + x_1 ) (1 + x_2 )... ( 1 + x_n ) \ge ( 1 + (x_1 x_2 ... x_n ) ^\frac {1} {n } ) ^ n $ for all $x_i > 0$.
I started by taking logarithm on both sides and trying to use the concavity of logarithm but this reverses the inequality. I don't really know how to start, any help is apprecaited.
 A: Note that the function $f(x) = \ln(1+e^x)$ is convex. Therefore,
\begin{align*}
\frac{1}{n}\sum_{i=1}^n \ln(1+x_i) &=\frac{1}{n}\sum_{i=1}^n \ln (1+e^{\ln x_i})\\
&\ge \ln\left(1+e^{\frac{1}{n}\sum_{i=1}^n \ln x_i}\right)\\
&=\ln\big(1+(x_1x_2\cdots x_n)^{1/n}\,\big).
\end{align*}
The required inequality follows immediately.
A: The general term for LHS is $$\pi(\sum(1+x_{i})$$ for eg $$(1+x_1)(1+x_2)=1+x_1x_2+x_1+x_2$$ for $$(1+x_1)..(1+x_3)=1+x_1x_2x_3+x_1x_3+x_2x_3+1$$ while rhs is $$\sum(π(1+^n \sqrt{x_i})^n)$$ so i assume $(x_1..x_n)^{1/n})=a$ so by binomial theorem its $\sum{n\choose k}.a^k=(x_1..x_n)+\frac{1}{((x_1..x_n)^{1/n})}^1..+1$ so it can be easily observed that $π(\sum(1+x_i)>=(x_1..x_n)+\frac{1}{((x_1..x_n)^{1/n})}^1$. 
A: by induction:
$n=2, (1+x_1)(1+x_2)=1+x_1+x_2+x_1x_2 \ge 1+2\sqrt{x_1x_2}+x_1x_2=(1+(x_1x_2)^{\frac{1}{2}})^2$
when $n=k, (1+x_1)(1+x_2)...(1+x_k) \ge (1+(x_1x_2...x_k)^{\frac{1}{k}})^k$ holds,
$n=k+1, (1+x_1)(1+x_2)...(1+x_k)(1+x_{k+1}) \ge (1+(x_1x_2...x_k)^{\frac{1}{k}})^k(1+x_{k+1}) $
now, we prove 
$(1+(x_1x_2...x_k)^{\frac{1}{k}})^k(1+x_{k+1}) \ge (1+(x_1x_2...x_kx_{k+1})^{\frac{1}{k+1}})^{k+1}$
let $a=(x_1x_2...x_k)^{\frac{1}{k}},x=x_{k+1} \implies (1+a)^k(1+x) \ge (1+(a^kx)^{\frac{1}{k+1}})^{k+1} \\ \iff \ln{(1+a)^k}+\ln{(1+x)} \ge (n+1) \ln{((1+(a^kx)^{\frac{1}{k+1}})} $
$f(x)=\ln{(1+a)^k}+\ln{(1+x)}-\ln{((1+(a^kx)^{\frac{1}{k+1}})}$
$f'(x)=\dfrac{1}{x}(\dfrac{x}{1+x}-\dfrac{y}{1+y}),y=(a^kx)^{\frac{1}{k+1}}$
$\dfrac{x}{1+x}$ is mono increasing function,$\implies \\ y=x(x=a), f'(x)=0, \\ y>x (x<a),f'(x)<0,\\ y<x(x>a),f'(x)>0 \implies \\f(x=a)=f_{min}=0 \implies f(x) \ge 0$
QED
