# AM-GM & Minimization Proof [duplicate]

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 $$\min\bigg(\cfrac{x+y}{2}\bigg) = \sqrt{xy}$$. $$\qquad \square$$

• No. The minimum can not be an "expression". It must be only a finite number. Indeed, you can also write $$x+y+1≥3\sqrt [3]{xy}\implies \frac {x+y}{2}≥\frac {3\sqrt [3]{xy}-1}{2}$$ Therefore, the minimum can not be considered as a non-constant expression. Jan 13 at 4:17
• Your claim about the minimum is false. With $x, y > 0$ the LHS has no minimum and gets arbitrarily close to $0$. Jan 13 at 4:37

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 $$y$$.
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.