How to prove that $\frac{a+b}{2} \geq \sqrt{ab}$ for $a,b>0$? I am reading a chapter about mathematical proofs. As an example there is:
Prove that: $$(1) \space\space\space\space\space\space\space\space\space\space\space \frac{a+b}{2} \geq \sqrt{ab}$$ for $a,b>0$. There is stated that the thesis for the proof is short multiplication formula: $$(2)\space\space\space\space\space\space\space\space\space(a+b)^2=a^2+2ab+b^2$$ Substracting from the short multiplication $(2)$ formula $4ab$  and using square root yields the proof by implication. (here ends the example in a book)
Please tell me if I am right about it: 


*

*we are considering only $a,b \in \Re^+$

*$(1)$ can be transformed into $a-2\sqrt{ab}+b \geq0$ which is an equivalent short multiplication formula $(x-y)^2=x^2-2xy+y^2$

*taking into account assumptions squared number is always equal or greater than zero.

*the implication says that if squared number is greater than zero then $(1)$ is true.


My first question is: is my thinking proper?
My other question is why the authors use $(a+b)^2$ and susbstract $4ab$ instead of starting with $(a-b)^2$. I am also confused about the symbols used - in $(1)$ and $(2)$ authors used $a$ and $b$ - is this just coincidence? Or this would be OK as well if in the first and second $a$ and $b$ were the same numbers?

$
\newcommand{\calc}{\begin{align} \quad &}
\newcommand{\calcop}[2]{\\ #1 \quad & \quad \text{"#2"} \\ \quad & }
\newcommand{\endcalc}{\end{align}}
$Here is an alternative proof.
Your (correct) proof has the problem that it starts with two 'rabbits' that are pulled out of a hat: first the fact that one should rewrite to get the expression $\;a-2\sqrt{ab}+b\;$, and then the fact that $\;a-2\sqrt{ab}+b \;=\; (\sqrt a- \sqrt b)^2\;$.  The earlier answers do something similar.
Personally, I prefer the following longer proof, which I find simpler, and which first gets rid of the awkward $\;\sqrt{\phantom{ab}}\;$ and then simplifies further:
$$\calc
\frac{a+b}{2} \geq \sqrt{ab}
\calcop{\equiv}{square both sides, $\;\geq\;$ unchanged because both sides are $\;\geq 0\;$}
\left(\frac{a+b}{2}\right)^2 \geq ab
\calcop{\equiv}{multiply both sides by $4$ -- to get rid of the fraction}
(a+b)^2 \geq 4ab
\calcop{\equiv}{expand square; simplify -- the only way to make progress}
a^2 - 2ab + b^2 \geq 0
\calcop{\tag{*} \equiv}{re-introduce square}
(a-b)^2 \geq 0
\calcop{\equiv}{every square is non-negative}
\text{true}
\endcalc$$
In this proof the only real insight that is required is in step $\text{(*)}$, and it is debatable how much of a rabbit it is.
