<s>[INST] 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?
 [/INST] if $\sqrt{a+b} = \sqrt a + \sqrt b$ for some $a,b \in \mathbb{R^{\geq 0}}$, then $a+b = a + 2\sqrt{a}\sqrt{b} + b$, then $\sqrt{a}\sqrt{b} = 0$. So $a =0$ or $b=0$.
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