$ x \ge 0\text{ and } y \ge 0 \implies \frac{x+y}{2} \ge \sqrt{xy} $

Question

The above applies $\forall x,y \in \mathbb{R}$

I've tried: $x + y \ge 0$

$$x + y \ge x$$

$$ (x + y)^2 \ge 2xy$$

$$\frac{(x + y)^2}{2} \ge xy$$

But the closest I get is $\dfrac{x+y}{\sqrt{2}} \ge \sqrt{xy}$

Any ideas?

For what it's worth, you are pretty close. You need to show a slightly(?) stronger inequality, $(x+y)^2 \ge 4xy$. — hardmath, Feb 04 '16 at 01:07
Several questions about the same inequality: http://math.stackexchange.com/questions/64881/proving-the-am-gm-inequality-for-2-numbers http://math.stackexchange.com/questions/904827/how-to-prove-that-fracab2-geq-sqrtab-for-a-b0 http://math.stackexchange.com/questions/1114615/if-0ab-prove-that-a-sqrtab-fracab2b http://math.stackexchange.com/questions/1150895/let-a0-and-b0-prove-that-sqrtab-le-ab-2 http://math.stackexchange.com/questions/1632763/how-to-go-upon-proving-fracxy2-ge-sqrtxy — Martin Sleziak, Dec 24 '16 at 20:08

score 5 · Answer 1 · answered Feb 04 '16 at 00:44

5

Note that: $$(x-y)^2\ge 0\implies x^2+y^2\ge2xy\implies x^2+2xy+y^2\ge4xy\implies(x+y)^2\ge4xy$$

answered Feb 04 '16 at 00:44

AsdrubalBeltran

score 3 · Accepted Answer · answered Feb 04 '16 at 00:45

3

$$(x-y)^2 \ge 0$$ $$x^2 - 2xy + y^2 \ge 0 $$ $$x^2 + y^2 \ge 2xy $$ $$x^2 + 2xy + y^2 \ge 4xy $$ $$(x+y)^2 \ge 4xy $$

answered Feb 04 '16 at 00:45

L.F. Cavenaghi

2 Answers2