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?

Write it as $,x-2\sqrt{x}\sqrt{y}+y \ge 0,$ and recognize a different square on the LHS. — dxiv, Aug 26 '18 at 19:41
@dxiv how did you achieve that expression? Is it possible to square root the entire right side and achieve that? I thought you couldn't simplify the root of the right side. — vik1245, Aug 26 '18 at 19:44
@BobSmith $\sqrt{xy} ≤ \frac{x+y}{2} \iff 2\sqrt{xy} \le x+y \iff 2\sqrt{xy} \color{red}{- 2 \sqrt{xy}}\le x+y\color{red}{- 2 \sqrt{xy}},$ — dxiv, Aug 26 '18 at 19:46

score 3 · Accepted Answer · answered Aug 26 '18 at 19:50

3

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

answered Aug 26 '18 at 19:50

score 2 · Answer 2 · answered Aug 26 '18 at 19:50

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

answered Aug 26 '18 at 19:50

valer

score 2 · Answer 3 · answered Aug 26 '18 at 19:51

2

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.

answered Aug 26 '18 at 19:51

3 Answers3