Check and comment my proof of $a+b \geq 2 \sqrt{ab}$ 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 $b\geq 0$.

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.
