How to go upon proving $\frac{x+y}2 \ge \sqrt{xy}$? I'm trying to prove this but am having some difficulty. 
For any $x,y\in\mathbb R$ such that $x\ge 0$ and $y\ge 0$ we have
$$\frac{x+y}2 \ge \sqrt{xy}.$$

So far what I have gotten to is $\frac{x+y}{2} \geq \frac{\sqrt{xy}}{2} $
After this point I don't know what to do. To get to this point:
$$y \geq 0 $$
$$\implies y \geq y$$
$$\implies xy \geq xy$$
$$\implies 2xy \geq xy$$
$$\implies x^{2} + 2xy + y^{2} \geq xy$$
$$\implies (x+y)^{2} \geq xy$$
Sqrt both sides to get:
$$\implies x+y \geq \sqrt{xy}$$
$$\implies\frac{x+y}{2} \geq \frac{\sqrt{xy}}{2}$$
 A: $$(x-y)^2 \ge 0\\ x^2+2xy+y^2-4xy \ge 0 \\(x+y)^2 \ge 4xy\\\frac {x+y}2 \ge \sqrt{xy}$$
A: Let us start from the beginning and simplify what we want to show: 
$$\frac{x+y}{2}\geq \sqrt{xy}$$
By multiplying both side by $2$, we will get:
$$x+y\geq 2\sqrt{xy}$$
Now, subtracting $2\sqrt{xy}$ from both sides, we left with:
$$x-2\sqrt{xy}+y\geq 0$$
Now, a question:
Why $x-2\sqrt{xy}-y$ is indeed greater(or equal) than zero?
Closer look on the above expression, one might see that its equal to $(\sqrt{x}-\sqrt{y})^2$
So, if we are given two positive real numbers $x$ and $y$, then without loss of generality we can say that $x>y$(why?), therefore $\sqrt{x}>\sqrt{y}$(prove that).
Hence:
$$\sqrt{x}-\sqrt{y}>0$$
And...
A: This is a comment on your attempt to prove this inequality rather than answer to your question. Just to show that it was not completely dead end, as you wrote in your comment.
You wrote that you tried this:

So far what I have gotten to is $\frac{x+y}{2} \geq \frac{\sqrt{xy}}{2} $
After this point I don't know what to do. To get to this point:
$$y \geq 0 $$
  $$\implies y \geq y$$
  $$\implies xy \geq xy$$
  $$\implies 2xy \geq xy$$
  $$\implies x^{2} + 2xy + y^{2} \geq xy$$
  $$\implies (x+y)^{2} \geq xy$$
Sqrt both sides to get:
$$\implies x+y \geq \sqrt{xy}$$
  $$\implies\frac{x+y}{2} \geq \frac{\sqrt{xy}}{2}$$


What you could do know could be ask yourself: Well, I want to prove stronger inequality. Could I get it by similar approach?
So we start from the end. You want to get
$$\frac{x+y}{2} \geq \sqrt{xy}$$
which is the same as 
$$x+y \geq 2\sqrt{xy}.$$
Relation between these two is the same as in the last step of your attempt - just divided by $2$. If we try to go one step back in the same way as above, this can be obtained as the square root of
$$
\begin{align*}
(x+y)^2 &\ge 4xy\\
x^2+2xy+y^2 &\ge 4xy\\
\end{align*}
$$
So at this point you could think about: "How could I get $x^2+2xy+y^2 \ge 4xy$?" Maybe after playing a bit with this expression you would found a way to prove this. (However, now you already know how to show this from other answers.)
