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.
Holder's inequality for sums states that for $p, q \in (1, \infty)$ with $\frac{1}{p}+\frac{1}{q}=1$, $$\sum_{k=1}^{n}{|a_kb_k|} \leq \left(\sum_{k=1}^{n}{|a_k|^p} \right)^{\frac{1}{p}}\left(\sum_{k=1}^{n}{|b_k|^q} \right)^{\frac{1}{q}}$$
(For $p=q=2$ this reduces to Cauchy Schwarz)
Applying this for $p=\frac{4}{3}, q=4$ (Note $a, b, c, d>0$) gives
$$(a^3+b^3) \leq \left( (a^3c^{\frac{3}{4}})^{\frac{4}{3}}+(b^3d^{\frac{3}{4}})^{\frac{4}{3}} \right)^{\frac{3}{4}}\left( \left(\frac{1}{c^{\frac{3}{4}}}\right)^4+\left(\frac{1}{d^{\frac{3}{4}}}\right)^4 \right)^{\frac{1}{4}}$$
$$(a^3+b^3)^4 \leq (a^4c+b^4d)^3 \left( \frac{1}{c^3}+\frac{1}{d^3} \right)=(a^4c+b^4d)^3\frac{c^3+d^3}{c^3d^3}$$
Thus
$$a^4c+b^4d \geq \left(\frac{(a^3+b^3)^4c^3d^3}{c^3+d^3} \right)^{\frac{1}{3}}=cd$$
$\frac{A^4}{d} >= 1-\frac{B^4}{c}$
$\frac{A^4}{d} >= \frac{(c-B^4)}{c}$
$\frac{C\times A^4}{d} >= C-B^4$
$C\times A^4 >= CD -D\times B^4$
$C\times A^4 + D\times B^4 >= CD$
