Proving:$(1+x)(1+x^2)(1+x^3)\cdots(1+x^n)\ge(1+x^{\frac{n+1}{2}})^n$ How to prove that :
$$(1+x)(1+x^2)(1+x^3)\cdots (1+x^n)\ge(1+x^{\frac{n+1}{2}})^n$$
 A: $$(1+x^r)(1+x^{n+1-r})=1+x^r+x^{n+1-r}+x^{n+1}\ge 1+x^{n+1}+2\sqrt{x^r\cdot x^{n+1-r}}$$ applying A.M.$\ge$ G.M. with $x^r,x^{n+1-r}$ assuming $x>0$ 
So, $$(1+x^r)(1+x^{n+1-r})\ge(1+x^{\frac{n+1}2})^2 $$
Now, $1\le r<n+1-r\implies r<\frac{n+1}2$
If $n$ is odd $=2m+1$ (say), $r<\frac{2m+1+1}2\implies 1\le r\le m=\frac{n-1}2$,
 so $n+1-r=2m+1+1-r$ will vary in $[m+2=\frac {n+3}2,n]$
 so $1+x^{\frac{n+1}2}$ will remain unpaired.
So, $$\prod_{1\le r\le n}(1+x^r)=(1+x^{\frac{n+1}2})\prod_{1\le r\le\frac{n-1}2}(1+x^r)(1+x^{n+1-r})\ge (1+x^{\frac{n+1}2})(1+x^{\frac{n+1}2})^{\left(2\cdot\frac{n-1}2\right)}=(1+x^{\frac{n+1}2})^n$$
Similarly if $n$ is even $=2m$ (say), $1\le  r<\frac{2m+1}2\implies 1\le r\le m=\frac n2,$
so $n+1-r=2m+1-r$ will vary in $[m+1=\frac n2+1,n]$
$$\prod_{1\le r\le n}(1+x^r)=\prod_{1\le r\le\frac n2}(1+x^r)(1+x^{n+1-r})\ge (1+x^{\frac{n+1}2})^{\left(2\cdot\frac n2\right)}=(1+x^{\frac{n+1}2})^n$$
A: It's wrong of course! Try $n=3$ and $x=-1$.
For non-negative $x$ it's just Holder:
$$(1+x)(1+x^2)...(1+x^n)\geq\left(1+\sqrt[n]{x\cdot x^2\cdot...\cdot x^n}\right)^n=\left(1+x^{\frac{n+1}{2}}\right)^n$$
