<s>[INST] AM-GM & Minimization Proof 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 $\min\bigg(\cfrac{x+y}{2}\bigg) = \sqrt{xy}$. $\qquad \square$
 [/INST] 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 $y$.
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