# A $4$ variable inequality [closed]

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.

Because by Holder $$\left(\frac{a^3}{c}+\frac{b^3}{d}\right)^2(c^2+d^2)\geq(a^2+b^2)^3$$ and we are done!