A $4$ variable inequality If $a,b,c,d$ are positive numbers such that $c^2+d^2=(a^2+b^2)^3$, prove that 
$$\frac{a^3}{c} + \frac{b^3}{d} \ge 1,$$
with equality if and only if $ad=bc$.
Source: Don Sokolowsky, Crux Mathematicorum, Vol. 6, No. 8, October 1980, p.259.

Cauchy-Schwarz gives 
$$\left( \frac{a^3}{c} + \frac{b^3}{d}\right)(ac+bd)\ge (a^2+b^2)^2.$$
So it suffices to show that 
$$(a^2+b^2)^2 \ge ac + bd.$$
Squaring, this is equivalent to showing 
$$(a^2+b^2)^3(a^2+b^2) \ge(ac+bd).$$
This follows immediately from the hypothesis that $(a^2+b^2)^3=c^2+d^2$ and Cauchy-Schwarz. 
