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?

$$\alpha(\beta+1)\leq\frac{5}{3}\alpha^2+\frac{1}{3}\beta^2$$

$$\frac{5}{3}\alpha^2+\frac{1}{3}\beta^2-\alpha(\beta+1) \geq 0$$

$$5\alpha^2+\beta^2-3\alpha\beta-3\alpha \geq 0$$

Solution 1:

Your required inequality can be adjusted as follows:

$$\begin{equation}\begin{aligned} 0 & \le 5\alpha^2 + \beta^2 - 3\alpha\beta - 3\alpha \\ & = 4\alpha^2 - 4\alpha\beta + \beta^2 + \alpha^2 + \alpha\beta - 3\alpha \\ & = (2\alpha - \beta)^2 + \alpha(\alpha + \beta - 3) \end{aligned}\end{equation}\tag{1}\label{eq1A}$$

Since $(2\alpha - \beta)^2 \ge 0$ and $\alpha \ge 0$, then \eqref{eq1A} can only possibly not be true if

$$\alpha + \beta - 3 \lt 0 \implies \alpha + \beta \lt 3 \tag{2}\label{eq2A}$$

If $\alpha = 0$, then \eqref{eq1A} is always true. If $\alpha = 1$, then \eqref{eq2A} means either $\beta = 0$, which gives the RHS of \eqref{eq1A} to be $2$, or $\beta = 1$, which gives the RHS of \eqref{eq1A} to be $0$. Next, for $\alpha = 2$, \eqref{eq2A} shows only $\beta = 0$ is possible, with this giving the RHS of \eqref{eq1A} to be $2$. Finally, for $\alpha \ge 3$, \eqref{eq2A} shows there's no non-negative value for $\beta$ possible.

This confirms \eqref{eq1A} is always true for all non-negative integer values of $\alpha$ and $\beta$.