<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] Let $a,b \in \mathbb R^+$ and suppose for the sake of contradiction that $\sqrt{a+b}=\sqrt{a}+\sqrt{b}$. Squaring both sides we get $$a+b = a+b+2\sqrt{ab}$$
$$\implies 2\sqrt{ab}=0 \implies ab=0$$ a contradiction as $a,b \neq 0$. Therefore $\sqrt{a+b} \neq \sqrt{a}+\sqrt{b}$ anytime $a$ and $b$ are nonzero.  
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