# Cauchy Schwartz Inequality Question: $(a^2+b^2)^3=c^2+d^2 \implies \frac{a^3}{c}+\frac{b^3}{d}\geq 1$

If $$(a^2+b^2)^3=c^2+d^2$$, prove that $$\frac{a^3}{c}+\frac{b^3}{d}\geq 1$$.

Hint: Use $$(a^{2} + b^{2})(c^{2} + d^{2}) \geq (ac + bd)^{2}$$ and $$\left( \frac{a^{3}}{c} + \frac{b^{3}}{d}\right)(ac+bd) \geq (a^{2} + b^{2})^{2}.$$
Because by Holder $$\left(\frac{a^3}{c}+\frac{b^3}{d}\right)^2(c^2+d^2)\geq(a^2+b^2)^3.$$
Using the standard and the Angel forms of the Cauchy-Schwarz inequality: $$LHS = \dfrac{(a^2)^2}{ac}+ \dfrac{(b^2)^2}{bd}\ge \dfrac{(a^2+b^2)^2}{ac+bd}= \dfrac{(a^2+b^2)^3}{(a^2+b^2)(ac+bd)}= \dfrac{c^2+d^2}{(a^2+b^2)(ac+bd)}\ge \dfrac{c^2+d^2}{(a^2+b^2)\cdot \sqrt{a^2+b^2}\cdot \sqrt{c^2+d^2}}= \sqrt{\dfrac{c^2+d^2}{(a^2+b^2)^3}}= \sqrt{1} = 1= RHS.$$