Analysis Homework. Proof of Arithmetic-Means Inequality

[The_Iceflash]

From the Text, Introduction to Analysis, by Arthur Mattuck pg 32 2-3 (a)

Homework Statement

Prove: for any a,b $$\geq$$ 0, $$\sqrt{ab}$$ $$\leq$$ $$\frac{\left(a+b\right)}{2}$$
with equality holding if and only if a =b

Homework Equations

All Perfect squares are $$\geq$$ 0

The Attempt at a Solution

I wasn't sure where to go with this but I took the inequality:
$$\sqrt{ab}$$ $$\leq$$ $$\frac{\left(a+b\right)}{2}$$

and squared both sides to give me:

ab $$\leq$$ $$\frac{\left(a+b\right)^{2}}{4}$$

I then separated the right which gave me:

ab $$\leq$$ $$\frac{a^{2}}{4}$$ + $$\frac{ab}{2}$$ + $$\frac{b^{2}}{4}$$

I multiplied both sides by 4 which gave me:

4ab $$\leq$$ $$a^{2}$$ + $$2ab$$ + $$b^{2}$$

I quickly saw that 4ab $$\leq$$ $$\left(a+b\right)^{2}$$

From the known equation: 0 $$\leq$$ $$\left(a+b\right)^{2}$$

I added the two together and got: 4ab $$\leq$$2 $$\left(a+b\right)^{2}$$

which divided by 4 equals: ab $$\leq$$ $$\frac{\left(a+b\right)^{2}}{2}$$

This as far as I got. Help would be appreciated.

 

[Mark44]

See this thread: https://www.physicsforums.com/showthread.php?t=376036

 

[The_Iceflash]

See this thread: https://www.physicsforums.com/showthread.php?t=376036

Thanks.