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Define $f:[0,\infty)^2\to\mathbb{R}$ as $f(x,y)=\sqrt{x}-\sqrt{x+y}$. I'm trying to see if $f$ is concave, i.e., if for every $t\in[0,1]$ and $(x_1,y_1),(x_2,y_2)\in[0,\infty)^2$, $$ f(tx_1+(1-t)x_2,ty_1+(1-t)y_2)\geq tf(x_1,y_1)+(1-t)f(x_2,y_2). $$ I'm not sure if it is true. One inequality that can help is $-\sqrt{y}\leq f(x,y)\leq \sqrt{y}$, obtained via triangle inequality.

Can anyone provide some hint?

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Consider the function $f(0,y) = -\sqrt{y}$.

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  • $\begingroup$ Thanks! That is indeed an easy explanation of why it is not concave. Do you think that the answer is still no if instead we define $f$ on $(0,\infty)^2$? $\endgroup$
    – RLC
    Apr 27, 2020 at 0:07
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    $\begingroup$ We can still fix $x$ (in any domain you choose) and look at the resulting strictly convex function of $y$, so the answer is still no. $\endgroup$
    – Brian Moehring
    Apr 27, 2020 at 0:10

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