How can I prove that $\sqrt{xy}$ is concave, not strictly concave? I tried to derivative twice $f''(x)$, but it becomes negative which is the definition for a strictly concave.
2026-03-30 10:44:17.1774867457
concave, not strictly concave
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$$f(x,y)=\sqrt{xy}=\sqrt{x}\sqrt{y}$$ $$\frac{\partial^2}{\partial_x\partial_y}f(x,y)=\frac{1}{4\sqrt{xy}}$$ Is this what you tried?