This is in connection to a problem in Economics. I am trying to draw the graph of $U(x,y)=\min(x+y,2\sqrt{xy})$. In my attempt I tried break the definition into different cases.
Now , $x+y=2\sqrt{xy} \implies x+y-2\sqrt{xy}=0\implies(\sqrt{x}-\sqrt{y})^2=0$
If $x+y\leq2\sqrt{xy}$, $U(x,y)=x+y$
That is, if $(\sqrt{x}-\sqrt{y})^2\leq0$, $U(x,y)=x+y$
If $x+y>2\sqrt{xy}$, $U(x,y)=2\sqrt{xy}$
That is, if $(\sqrt{x}-\sqrt{y})^2>0$, $U(x,y)=2\sqrt{xy}$
I am stuck here. How do I draw the graph of this function?
I assume you're looking at this for $x, y \ge 0$, or else $2\sqrt{xy}$ wouldn't be well-defined. In that case, $\sqrt{x}$ and $\sqrt{y}$ make sense, and as you observe, the two "parts" of the function are equal exactly when $$ (\sqrt{x} - \sqrt{y})^2 = 0. $$ But this implies that $$ \sqrt{x} - \sqrt{y} = 0\\ \sqrt{x} = \sqrt{y}\\ x = y $$ That leaves two cases: $x > y$ and $x < y$. We want to know which of the two parts of $U$ is larger on each piece. Fortunately, if we look at $U/2$, we see that it's the smaller of $\frac{x+y}{2}$ and $\sqrt{xy}$, the first of which is the "arithmetic mean" and the second is the "geometric mean". There's a classic theorem (the arithmetic-geometric inequality) that says that the geometric mean is never larger than the arithmetic one, i.e., $\sqrt{xy} \le \frac{x+y}{2}$ (for positive $x, y$). So we can rewrite $$ U(x, y) = \min (x + y, 2\sqrt{xy}) = 2 \sqrt{xy}. $$ And that is pretty easy to graph.