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 

Using Titu's Lemma, we have 
$$ \dfrac{a^3}{c} + \dfrac{b^3}{d} \ge \dfrac{(a^2+b^2)^2}{ac+bd}$$
So, we are left to prove that
$$ (a^2+b^2)^2 \geq (ac+bd)\tag{1} $$
Using Cauchy-Schwarz inequality, we have 
$$(a^2+b^2)(c^2+d^2) \geq (ac+bd)^2\tag{2}$$
Using the given proposition, $$c^2+d^2 =(a^2+b^2)^3$$ in $(2)$ and the fact that $a,b,c,d \in \mathbb{R^+}$, we get $(1)$, which was required to be proved.
