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?

a,b positive real numbers.
Let $x^2:= a$, $y^2:=b,$  $x,y \gt 0.$
Consider a right triangle with lengths of legs $x,y$,
length of hypotenuse is $(x^2+y^2)^{1/2}.$
Sum of $2$ sides in a triangle is greater than the third side.
(Euclid)
$\Rightarrow:$
$(x^2+y^2)^{1/2} \lt x + y$, 
or reverting to $a,b:$
$(a+b)^{1/2} \lt a^{1/2}+b^{1/2}.$
