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?
 A: 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.
A: This proof is hinted by the presence of the square root, which one will tend to remove by squaring. As all numbers are positive
$$\sqrt{xy}\le\frac{x+y}2$$ is rewritten
$$xy\le\frac{x^2+2xy+y^2}4,$$
which is also 
$$0\le\frac{x^2-2xy+y^2}4$$ and certainly holds (see why ?).

Once you have understood this principle, you can recast the proof as
$$\sqrt x\sqrt y\le\frac{\sqrt x^2+\sqrt y^2}2$$ or $$0\le(\sqrt x-\sqrt y)^2.$$

An alternative way to eliminate the square root is to set, like in Narasimham's answer, $x=u^2,y=v^2$ and prove
$$uv\le\frac{u^2+v^2}2.$$
A: Let $ x=u^2, y=v^2$ since $ (x,y)$ are given positive.
You have to prove that
$$ u^2+v^2-2 u v \ge 0$$
or that
$$(u-v)^2\ge 0 $$
which is true for all real numbers.
A: $$\sqrt(xy) \leq \frac{x+y}{2}$$
$$xy \leq \frac{(x+y)^2}{4}$$
$$4xy \leq x^2+2xy+y^2$$
$$2xy \leq x^2+y^2$$
$$2 \leq \frac{x^2}{xy}+\frac{y^2}{xy}$$
$$2 \leq \frac{x}{y}+\frac{y}{x}$$
Since $\frac{x}{y}$ is a fraction, assume $x$ is greater than $y$, say
$$z = \frac{x}{y}$$
This means that the lowest value of $z$ is $1$, if $x > y$
$$2 \leq z + \frac{1}{z}$$
If $z$ is lowest, $\frac{1}{z}$ would be highest, but the minimum value of their sum comes when they're minimum.....equal to $1$
$$z + \frac{1}{z} = ≥ 2$$
A: The well-known fact your teacher is talking about is that the square of any really number is nonnegative.  That is a theorem.  As for why you should start with this particular instance of that theorem, that takes a little insight.  
If you were trying to prove the theorem yourself, you  might start by seeing what happens if it isn't true.  $$\sqrt{xy}>\frac{x+y}2\\
xy>\frac{x^2+2xy+y^2}4\\
4xy>x^2+2xy+y^2\\
0>x^2-2xy+y^2\\
0>(x-y)^2$$
contradiction.
This is a perfectly good proof, but you might prefer a direct proof.  If you try to work backwards, you immediately run into trouble, because taking the square root of both sides may not be valid.  It may then occur to you to start with the square roots.   
