TITLE: How prove this inequality $2a^ab^bc^cd^d\ge ac+bd$
QUESTION [20 upvotes]: Let $a,b,c,d$ be positive numbers such that $a+b+c+d=2$. Show that
$$2a^ab^bc^cd^d\ge ac+bd$$
My try: I think maybe I can use this inequality
$$(1+x)^n\ge 1+nx \hspace{12pt} (n>1)$$
then I can't get it to work.
I think this inequality may be solved by nice methods. Thank you.

REPLY [2 votes]: This is only for proof of mercio's last inequality:
$f(x)=x\log{(1+x)}-x\log{(1-x)}+\log{(1-x^2)}-\log{(1+x^2)}$,
since $f(x)=f(-x)$, we only take $x\ge0$
now prove $f'(x)>0$:
$f'(x)=\log{\dfrac{1+x}{1-x}}-\dfrac{2x}{x^2+1}$
let $g(x)=\log{\dfrac{1+x}{1-x}}-2x, g'(x)=\dfrac{2x^2}{1-x^2}>0 \implies g(x)>0 \implies \log{\dfrac{1+x}{1-x}}>2x\ge\dfrac{2x}{x^2+1} \implies f'(x)>0 \implies f(x)\ge f(0)=0 $