Question on basic Inequality of arithmetic and geometric means

Question

It appears that I have made some kind of error, considering that I have arrived at a contradiction. Can someone point out where things got derailed?

"Prove $\frac{a+b}{2} \geq \sqrt{ab} \text{ for } 0<a\leq b$"

$(a+b)^2 \geq (2\sqrt{ab})^2 \rightarrow $ $a^2+2ab+b^2 \geq 4ab \rightarrow a^2-2ab+b^2 \geq 0 \rightarrow$ $(a-b)^2 \geq 0 \rightarrow a-b \geq 0 \rightarrow a \geq b \Rightarrow\Leftarrow\quad$

Answer 1

Proving AM-GM inequality for two numbers can be done in two different ways:

1. Arithmetically: The idea here is to start from the back(I can explain this further if you want). So you know that $(a - b)^2 ≥ 0 $ From there you can work out that $a^2 + b^2 ≥ 2ab$ If we add $2ab$ to both sides we get $(a + b)^2 ≥ 4ab \Rightarrow a+b ≥ 2\sqrt{ab} \Rightarrow \frac{a+b}{2} ≥ \sqrt{ab}$.
2. Geometrically: Let $ABC$($\angle C = 90^\circ$) be a right angled triangle. Let $CD$ be its height and $CE$ its median line. If $AD = p$ and $BD = q\Rightarrow CD = \sqrt{pq},CE = \frac{p+q}{2}$. Because CE is the hypotenuse of triangle $CDE$ we can say $CE ≥CD$ or $\frac{p+q}{2}≥\sqrt{pq}$

Answer 2

Square of a real number is always positive. You are done when you reach $(a-b)^2\ge0$ as it is obvious.