<s>[INST] Prove that $\frac{a^3}{b} + \frac{b^3}{a} \geqslant a^2 + b^2$ 
Prove that $$\frac{a^3}{b} + \frac{b^3}{a} \geqslant a^2 + b^2$$
for $a,b \in \mathbb{R}^+$. Establish when the equality holds.

My approach was to use AM-GM however trying both sides individually assert to $\geqslant 2ab.$ RHS is baby AM-GM if considered.
Edit:
From L.H.S -
$\dfrac{\frac{a^3}{b} + \frac{b^3}{a}}{2} \geqslant ab \implies \dfrac{a^3}{b} + \dfrac{b^3}{a} \geqslant 2ab$
 [/INST] Here is a trick, using only AM-GM, twice.
$$a^2+b^2\geq 2ab \Rightarrow \\
\frac{a^3}{b}+\color{red}{a^2}+\frac{b^3}{a}+\color{red}{b^2} \geq 
\frac{a^3}{b}+\color{red}{ab}+\frac{b^3}{a}+\color{red}{ab}\geq\\
2\sqrt{\frac{a^3}{b}\cdot ab} + 2\sqrt{\frac{b^3}{a}\cdot ab}=2a^2+2b^2$$
Or
$$\frac{a^3}{b}+\color{blue}{a^2}+\frac{b^3}{a}+\color{blue}{b^2} \geq 2(\color{blue}{a^2+b^2})$$
and the result follows.
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