Proof by Deduction $\sqrt{xy} ≤ \frac{x+y}{2}$ I want to ask a question about proof of deduction. 
I sat my Pure Mathematics Exam more than $3$ years ago but decided to return to the subject for a refresher. 
Proofs were not a requirement for my course but as my younger siblings are studying it, I decided to give it a go. 
This was the question:

Prove that for all positive values of x and y
  $$\sqrt{xy} ≤ \frac{x+y}{2}$$

Now, I did some research on proofs of deduction and it involved a start point. 
My instinct was to work backwards from this inequality to something more meaningful towards this "start point" and work forwards. 
This is my working thus far:
$$xy ≤ \frac{(x+y)^2}{4}$$
$$4xy ≤ (x+y)^2$$
$$4xy ≤ x^2 + 2xy + y^2$$
Unfortunately, I can't seem to see where I can go further to start this proof. 
Is this the correct approach? If so, is there a further step that I cannot see?

From your last step, you could proceed as follows:
$$
4xy \leq x^2 + 2xy + y^2\\
0 \leq x^2 - 2xy + y^2 = (x-y)^2.
$$
Now you can work backward as you wanted.
