Proof by Deduction $\sqrt{xy} ≤ \frac{x+y}{2}$

Question

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?

Answer 1

... $\rightarrow$ $0\le x^2-2xy+y^2=(x-y)^2$

Answer 2

$\sqrt{xy} ≤ \frac{x+y}{2} \iff 2\sqrt{xy} ≤ {x+y} \iff x-2\sqrt{xy}+y \geq 0$ $ \iff (\sqrt{x}-\sqrt{y})^2 \geq 0 $.
Which is true

Answer 3

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.