If positive $a,b,c,d$ satisfy $(a^3+b^3)^4=c^3+d^3$, prove that $a^4c+b^4d\ge cd$. 
If positive $a,b,c,d$ satisfy $(a^3+b^3)^4=c^3+d^3$, prove that $$a^4c+b^4d\ge cd$$

It kind of seems useful to begin with a division of both sides by $cd$:
$$\frac{a^4}{d}+\frac{b^4}{c}\ge1$$
It seems like a simple Cauchy-Schwarz would be of use in this case. However, after some attempts I can't come up with the right solution. It'd be great to hear some ideas. Thanks.

$\frac{A^4}{d} >= 1-\frac{B^4}{c}$
$\frac{A^4}{d} >= \frac{(c-B^4)}{c}$
$\frac{C\times A^4}{d} >= C-B^4$
$C\times A^4 >= CD -D\times B^4$
$C\times A^4 + D\times B^4 >= CD$
