<s>[INST] Showing $\alpha(\beta+1)\leq\frac{5}{3}\alpha^2+\frac{1}{3}\beta^2$ for nonnegative integers $\alpha$ and $\beta$. I found this lemma in a paper I was reading and it was not proved. There doesn't seem to be any obvious factorized form so how would one go about proving the inequality holds?
$$\alpha(\beta+1)\leq\frac{5}{3}\alpha^2+\frac{1}{3}\beta^2$$
$$\frac{5}{3}\alpha^2+\frac{1}{3}\beta^2-\alpha(\beta+1) \geq 0$$
$$5\alpha^2+\beta^2-3\alpha\beta-3\alpha \geq 0$$
 [/INST] Alternative proof inspired by Michael Hardy:
$$5\alpha^2+\beta^2-3\alpha\beta-3\alpha \geqslant 0 \iff 20\alpha^2+4\beta^2-12\alpha 
\beta-12\alpha \geqslant 0$$
$$20\alpha^2+4\beta^2-12\alpha \beta-12\alpha 
= (2\beta-3\alpha )^2+11\alpha^2-12\alpha \\
=\begin{cases}
(2\beta-3\alpha )^2+\alpha (11\alpha-12)>0,  & \text{if $\alpha>1$} \\
(2\beta-3)^2-1 \geqslant 0, & \text{if $\alpha=1$} \\
4\beta^2 \geqslant 0 & \text{if $\alpha=0$}
\end{cases}
$$
$$
"=" \iff \alpha=1, 2\beta-3=\pm 1 \text{ or } \alpha=\beta=0\\ \iff (\alpha, \beta)\in \{(0,0), (1,1), (1,2)\}
$$
</s>