Finding $\min_{x,y\ge0} (x+y)$ such that $\sqrt{x}+\sqrt{y}\ge c$ for a given constant $c\ge0$

Question

For a given real number $c\ge0$, we want to solve the following optimization problem: \begin{align} \min_{x,y}\ &x+y\\ \text{subject to}\ \ \ &\sqrt{x}+\sqrt{y}\ge c\\ &x,y\ge 0 \end{align}

To solve it, I first want to show that in an optimal solution we must have $\sqrt{x}+\sqrt{y} = c$. I am not sure if this step is required to solve this optimization problem. Although it seems to be obvious, I could not prove this property! To prove it by contradiction, I tried to show that the following sets of constraints cannot be true at the same time: \begin{align} &\sqrt{x}+\sqrt{y} > \sqrt{a}+\sqrt{b}\\ &x+y < a + b\\ &x,y, a , b\ge 0 \end{align}

I also tried to use the inequality $\sqrt{xy}\le\frac{x+y}{2}$ but it did not work. I am not sure what is wrong!

Answer 1

$$x+y\geq\frac{1}{2}(\sqrt{x}+\sqrt{y})^2\geq\frac{c^2}{2}$$

Equality when $x=y=\frac{c^2}{4}$

To show the first inequality you can just open the brackets and apply AM-GM

Answer 2

By C-S $$x+y=\frac{1}{2}(1+1)(x+y)\geq\frac{1}{2}(\sqrt{1\cdot {x}}+\sqrt{1\cdot y})^2\geq\frac{c^2}{2}.$$ The equality occurs for $(1,1)||(\sqrt{x},\sqrt{y})$ and $\sqrt{x}+\sqrt{y}=c,$ which says that we got a minimal value.

Answer 3

Since $x,y,c\ge 0$, you can square the constraint inequality: $$\sqrt{x}+\sqrt{y}\ge c \iff x+y+2\sqrt{xy}\ge c^2.$$ Now use AM-GM: $$x+y\ge 2\sqrt{xy}$$ to get: $$2(x+y)=x+y+x+y\ge x+y+2\sqrt{xy}\ge c^2 \Rightarrow x+y\ge \frac{c^2}{2},$$ the equality occurs for $x=y=\frac{c^2}{4}$.