Convexity argument to maximize $\sqrt{\alpha \beta} + \sqrt{(1-\alpha)(1-\beta)}$ over $\alpha,\beta\in [0,1]$

Question

The function $$f(\alpha,\beta) = \sqrt{\alpha\beta} + \sqrt{(1-\alpha)(1-\beta)}$$ achieves its maximum on $[0,1]^2$ by the line $\alpha = \beta$. This can be shown analytically. However, noting that $\sqrt{\cdot}$ is concave, it feels like there should be a simple convexity argument to show it is maximized on $\alpha = \beta$. Is there?

Answer 1

Convexity is too much, this can be done by a simple application of AM-GM (inequality between arithmetic mean and geometric mean). Indeed $$\sqrt{\alpha\beta} + \sqrt{(1 - \alpha)(1 - \ beta)} \le \frac{\alpha + \beta}{2} + \frac{(1 - \alpha) + (1 - \beta)}{2} = 1$$ with equality whenever $\alpha = \beta$.

Answer 2

Let $G(x,y)=\sqrt{xy}$. Then maybe $\frac{G(\alpha,\beta)+G(1-\alpha,1-\beta)}{2}\leq G(\frac{\alpha+1-\alpha}{2},\frac{\beta+1-\beta}{2})$. Why not use $\sqrt{\alpha\beta}\leq \frac{\alpha+\beta}{2}$.

Answer 3

By C-S $$\sqrt{\alpha\beta}+\sqrt{(1-\alpha)(1-\beta)}\leq\sqrt{\alpha+1-\alpha)(\beta+1-\beta)}=1.$$ The equality occurs for example for $\alpha=\beta=\frac{1}{2},$ which says that $1$ is a maximal value.