Check and comment my proof of $a+b \geq 2 \sqrt{ab}$

Question

I want to prove

$$ a+b \geq 2 \sqrt{ab}, \quad \text{for} \quad a, b>0 \tag 1 $$

First question:

Isn't this wrong? Shouldn't it be $$ a+b \geq 2 \sqrt{ab}, \quad \text{for} \quad a, b\geq0 \tag 2 $$

Or $$ a+b > 2 \sqrt{ab}, \quad \text{for} \quad a, b>0 \tag 3 $$

And the proof:

I start backwards. \begin{align} a+b-2&\sqrt{ab}\geq 0 \tag 4 \\ (\sqrt a)^2+(\sqrt b)^2-2&\sqrt{ab}\geq 0 \tag 5 \\ (\sqrt a)^2+(\sqrt b)^2-2&\sqrt{a}\cdot \sqrt{b}\geq 0 \tag 6 \end{align} Let $x= \sqrt a$ and $y=\sqrt b$, so we see that \begin{align} x^2+y^2-2xy&\geq 0 \tag 7\\ (x-y)^2\geq 0 \tag 8 \end{align} So finally we have $$ (\sqrt a- \sqrt b)^2 \geq 0 \tag 9 $$

If $a=b=0$ we have $0\geq 0$ which is true for $0=0$ ($0>0$ is always false).

For $a>0$ and $b>0$ the square $(\sqrt a- \sqrt b)^2$ is always strictly positive.

And also, $a<0$ and $b<0$ is not valid because $\sqrt a$ and $\sqrt b$ are only defined for $a\geq 0$ and $b\geq 0$.

Answer 1

No, it's not wrong. It is simply not as strong as it could be. Your (2) is also right, but that doesn't make (1) wrong.

However, your (3) is wrong: if $a=b$, then $a+b=2\sqrt{ab}$.

Answer 2

I would just want to point out that in the third equation and the proof, \begin{align} \text{for }~ a ,~ b > 0 , \ ( \sqrt{a} - \sqrt{b} )^2 \end{align} is non-negative, since both of them could be equal.

Answer 3

You want to prove $a+b \geq 2 \sqrt{ab}, \quad \text{for} \quad a, b>0$.

You can use $a^2 > 0$ for all $a \in \mathbb{R}$ as an assumption and segue into (9), with the restriction that $a,b > 0$. Then, instead of working backwards, you're working forwards and the proof is nice.

Rewrite your proof so that it begins with the assumptions, then (9) becomes (7) with $x = \sqrt{a}$ and $y= \sqrt{b}$ which then becomes (4), and finally conclude your proof.