AM-GM & Minimization Proof

Question

I want to prove that for all $x, y > 0$, $$\cfrac{x+y}{2} \geq \sqrt{xy}$$ Particularly, I want to show that the minimum of $(x+y)/2$ is exactly $\sqrt{xy}$.

This is my attempt:

$\textbf{Proof}$ (Contradiction). Assume if $x, y > 0$, then $(x+y)/2 < \sqrt{xy}$. But \begin{align*} x+y &< 2\sqrt{xy} \\ x^2+2xy+y^2 &< 4xy \\ x^2-2xy+y^2 &< 0 \\ (x-y)^2 &< 0 \end{align*} gives us a contradiction since for all $x, y > 0$, we know $(x-y)^2 > 0$. Therefore there exists no positive reals $x$ and $y$ such that $(x+y)/2 < \sqrt{xy}$, or $\min\bigg(\cfrac{x+y}{2}\bigg) = \sqrt{xy}$. $\qquad \square$

Answer 1

Instead of talking about minimum, we may say that the inequality $$\frac{x+y}{2} \geq \sqrt{xy}$$ holds for any $x, y>0$ and the equality holds for $x=y.$

We may prove the inequality directly by starting with $$ \begin{aligned} & x+y-2 \sqrt{x y}=(\sqrt{x}-\sqrt{y})^2 \geqslant 0 \\ \Rightarrow \quad & x+y \geqslant 2 \sqrt{x y} \\ \Rightarrow \quad & \frac{x+y}{2} \geqslant \sqrt{x y} \end{aligned} $$

By the way, we can say that the minimum value $x+y-2\sqrt{xy} $ is $ 0$ for positive values of $x$ and $y$.

Answer 2

I think your statement is even incorrect. The quantity $\sqrt{xy}$ is not a constant. Also, it depends on the domain which may or may not yield a minimum. For example, over $[1,\infty)\times [1,\infty)$ then there is a minimum. But over $(1,\infty)\times (1,\infty)$ there is no minimum.