<s>[INST] 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.
 [/INST] 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
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