How to prove that $\frac{a+b}{2} \geq \sqrt{ab}$ for $a,b>0$? [duplicate]

Argument is fine as long as it is clear to you that your are using equivalence and not implication on your way down the bullets. That is for $a, b \in \Bbb R^{+}$

$$ \frac{a+b}{2} \geq \sqrt{ab} \iff a-2\sqrt{ab}+b \geq0 \iff (\sqrt a- \sqrt b)^2 \ge 0 $$

You must use "double arrows". Or "if and only if" connectives. Only then can you justify the leftmost statement from the correctness of the one in the right.

The rightmost equality makes sense due to two reasons. One due to the assumption you mentioned above regarding the non-negativity of a square. And two the square roots of $a$ and $b$ exist since both are positive.

You are right. This is a better way to prove this in my opinion. So I wouldnot worry about it.

As for the notation : Yes the formula would still hold if $a$ and $b$ were both the same number. Toverify just look at your proof. Substitue $a$ for $b$ and see what happens. you will notice the equality in $\geq$ holds. This is in fact a secondary idea of this result. The equality holds if and only if $a = b$. You should try to prove this. Decent exercise.

$ \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.

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.