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?

Your method is fine.
Note, however, that the other method, which goes$$(a+b)^2-4ab=(a-b)^2\ge 0$$at this stage requires no square roots, so this works for all integers. We have the implication $$(a+b)^2\ge 4ab$$ so that if both sides are non-negative (i.e. $4ab\ge 0$) $$\frac {a+b}2\ge\sqrt {ab}$$
Some people prefer to take roots at the last possible moment because the expressions tend to be simpler, there is a danger that conditions for the roots to exist can get forgotten if the roots are taken part way through, and by taking a positive square root it is possible to lose a negative solution to the original problem (just as in squaring it is possible to gain a solution).

It is worth knowing different ways of manipulating these expressions, because in more complex contexts different things work best. For example:
$$(a-b)^2+(b-c)^2+(c-a)^2\ge 0$$ can be expanded, and after dividing by $2$ and rearranging you get $$a^2+b^2+c^2\ge ab+bc+ca$$ You wouldn't normally want to write this as $x+y+z\ge$ some expression involving square roots.
