How would one prove that for all positive real $a$ and $b$, $\sqrt{a+b} \neq \sqrt{a} + \sqrt{b}$? I think that the best way to prove this would be to prove by contradiction. Am I right?
If so, can I just assume that $\sqrt{a+b} = \sqrt{a} + \sqrt{b}$ and say $a = 1$ and $b = 1$ and show that $\sqrt{1+1} \neq \sqrt{1} + \sqrt{1}$, or is that not enough?

If
$\sqrt{a + b} = \sqrt a + \sqrt b, \tag 1$
then, squaring,
$a + b = a + 2\sqrt a \sqrt b + b, \tag 2$
whence
$2\sqrt a \sqrt b = 0, \tag 3$
which forces
$a = 0 \vee b = 0. \tag 4$
Contradiction!
