Q:

Concavity of the difference of square roots

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?

A:

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

