prove inequation a,b,c,d $\in \mathbb{R}$  $a,b,c,d \gt 0$ and  $ c^2 +d^2=(a^2 +b^2)^3$ prove that  $$ \frac{a^3}{c} + \frac{b^3}{d} \ge 1$$
If I rewrite the inequation like $ \frac{a^3}{c} + \frac{b^3}{d} \ge \frac{c^2 +d^2}{(a^2 +b^2)^3}$ and manage to simplfy  it brings me nowhere. I try with 
Cauchy-Schwarz Inequality but still can not solve it.
I would very much appreciate it if anyone could help me 

Another way would be Holder's inequality which gives:
$$\left(\frac{a^3}c+\frac{b^3}d \right)^2(c^2+d^2)\geqslant (a^2+b^2)^3$$
