The inequality $a^cb^d(c+d)^{c+d}\le c^cd^d(a+b)^{c+d}$ 
The inequality $a^cb^d(c+d)^{c+d}\le c^cd^d(a+b)^{c+d}$  for  $a,b,c,d>0$

The inequality is equivalent to: 
$$\displaystyle\frac{a^cb^d}{(a+b)^{c+d}}\le\frac{c^cd^d}{(c+d)^{c+d}}$$
and the right side should be at least $\left(\frac{1}{2}\right)^{c+d}$ if the rearrangement inequality is true also for reals. 
So it would be sufficient if left side is below $\left(\frac{1}{2}\right)^{c+d}$, but this seems to be true that if only $a,b,c,d\gt1$. 
can you help 

Since $f(x)=x\ln x$ is a convex function, by Jensen we obtain:
$$\frac{a}{a+b}\left(\frac{c}{a}\ln\frac{c}{a}\right)+\frac{b}{a+b}\left(\frac{d}{b}\ln\frac{d}{b}\right)\geq\left(\frac{a}{a+b}\cdot\frac{c}{a}+\frac{b}{a+b}\cdot\frac{d}{b}\right)\ln\left(\frac{a}{a+b}\cdot\frac{c}{a}+\frac{b}{a+b}\cdot\frac{d}{b}\right)$$ or
$$\left(\frac{c}{a}\right)^c\left(\frac{d}{b}\right)^d\geq\left(\frac{c+d}{a+b}\right)^{c+d},$$
which is your inequality.
Done! 
