How prove this inequality $2a^ab^bc^cd^d\ge ac+bd$ 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.
 A: If $a \neq c$, then look at what happens if we replace both $a$ and $c$ with $x = (a+c)/2$ :
$x^2 \ge x^2 - (\frac {a-c}2)^2 = ac$.
Since the function $x \mapsto x \log x$ is convex (its second derivative is $1/x$ which is positive),
$\log(a^ac^c) = a \log a + c \log c \ge 2x \log x = \log (x^{2x})$.
Hence $2a^ab^bc^cd^d \ge 2x^xb^bx^xd^d$ , $x^2+bd \ge ac+bd$ , and we still have $x+b+x+d = a+b+c+d = 2$ so it is enough to prove the inequality when $a=c$ (and when $b=d$, for symmetry reasons).
So we are left with proving that if $x+y = 1$ then $2x^{2x}y^{2y} \ge x^2 + y^2$. Let $x = \frac {1+z}2$ and $y = \frac {1-z}2$. After simplifying, we have to prove that for $|z| < 1$, $(\frac{1+z}{1-z})^z \frac {1 - z^2}{1 + z^2} \ge 1$.
Taking logs, we have to show $z \log(1+z) - z \log(1-z) + \log(1-z^2) - \log(1+z^2) \ge 0$.
Using $\log$'s power series at $1$, this is 
$(2z^2 + 2z^4/3 + 2z^6/5 + 2z^8/7 + 2z^{10}/9 +\ldots) - (2z^2 + 2z^6/3 + 2z^{10}/5 + \ldots) \ge
(2z^2 + 2z^6/3 + 2z^{10}/5 + 2z^{14}/7 + 2z^{18}/9 + \dots) - (2z^2 + 2z^6/3 + 2z^{10}/5 + \ldots) = 0$
There probably is something smarter to do instead, but this works.
A: 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 $
