Prove $\sqrt{ xy} \leq \frac{x + y}{2}$ for all positive $x$ and $y$ First and foremost, I have already gone through the following posts:
Prove that, for all positive integers $x$ and $y$, $\sqrt{ xy} \leq \frac{x + y}{2}$
Proving the AM-GM inequality for 2 numbers $\sqrt{xy}\le\frac{x+y}2$
The reason why I open a new question is because I do not understand after reading the two posts.
Question: Prove that for any two positive numbers x and y, $\sqrt{ xy} \leq \frac{x + y}{2}$
According to my lecturer, he said that the question should begin with $(\sqrt{x}- \sqrt{y})^2 \geq 0$. Lecturer also said that this is from a "well-known" fact. Now, both posts also mentioned this exact same thing in the helpful answers.
My question is this - how and why do I know that I need to use $(\sqrt{x}- \sqrt{y})^2 \geq 0$? What "well-known" fact is this? Can't I simply just subtract $\sqrt{xy}$ to both side and conclude at $0 \leq {(x-y)}^2$? I do not know how this $(\sqrt{x}- \sqrt{y})^2 \geq 0$ come back and why it even appear.
Thanks in advance.
Edit: I am not looking for the direct answer to this question. I am looking for an answer on why $(\sqrt{x}- \sqrt{y})^2 \geq 0$ is even considered in the first place as the first step to this question. Is this from a mathematical theorem or axiom etc?

We would like to prove
$$
\frac{{x + y}}{2} \ge \sqrt {xy} 
$$
for all non-negative $x$, $y$. If this was true, then we would also have
$$
x + y \ge 2\sqrt {xy},
$$
$$
x - 2\sqrt {xy}  + y \ge 0,
$$
$$
\left( {\sqrt x  - \sqrt y } \right)^2  \ge 0.
$$
But this last inequality is certainly true for all non-negative $x$, $y$. This is because the square of any real number is non-negative. Now you can start with this "well-known fact" and do everything backwards to obtain $\frac{{x + y}}{2} \ge \sqrt {xy} $. This shows you how one might come up with the idea of starting with $\left( {\sqrt x  - \sqrt y } \right)^2  \ge 0$. This is a common proof technique in mathematics when you have a series of equivalent transformations that can be performed in both directions.
