Find max and min of $2\sqrt{x+y} - \sqrt x - \sqrt y$ Find $max$ and $min$ of $M=2\sqrt{x+y} - \sqrt{x} - \sqrt {y}$ with $x + y = constant$
I tried to call $c = \sqrt{x+y} \Rightarrow M= 2c - (\sqrt x+\sqrt {y})\ge (2-\sqrt{2})c$ ? 

Since $x+y$ is a constant, to maximize $2\sqrt{x+y}-\sqrt{x}-\sqrt{y}$, we need to minimize what we take away from $2\sqrt{x+y}$, so we need to minimize $\sqrt{x}+\sqrt{y}$.
Similarly, to minimize $2\sqrt{x+y}-\sqrt{x}-\sqrt{y}$, we need to maximize $\sqrt{x}+\sqrt{y}$.
So let us first minimize $\sqrt{x}+\sqrt{y}$. This is equivalent to minimizing its square, which is $x+y+2\sqrt{x}\sqrt{y}$. But $x+y$ is fixed. What is the minimum possible value of $2\sqrt{x}\sqrt{y}$? 
Next we maximize $\sqrt{x}+\sqrt{y}$. Equivalently, we maximize its square $x+y+2\sqrt{x}\sqrt{y}$. So we want to maximize $4xy$. 
Note that $(x+y)^2=(x-y)^2 +4xy$. So $4xy=(x+y)^2-(x-y)^2$. Given that $x+y$ is fixed, what does this say about the maximum value of $4xy$?    
