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: 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!
A: Supose that there is an equality and that $ab\neq 0$. Then square both sides. Then conclude when you obtain a contradiction. 
A: Assume that exist $a,b>0$ such that $\sqrt{a+b}=\sqrt{a}+\sqrt{b}$
Then $(a+b)=a+b+2\sqrt{ab} \Rightarrow \sqrt{ab}=0 \Rightarrow a=0 \text{  or  } b=0$ which is a contradiction.
Because both $a,b$ are assumed to be positive.
A: 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 $b=0$.
A: 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.  
A: Yes you can!
If it's true then it should be true for 
$$a=b=1,$$
which gives $$\sqrt2=2,$$
which is a contradiction. 
A: 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 $a,b:$
$(a+b)^{1/2} \lt a^{1/2}+b^{1/2}.$
