How would one prove that for all positive real $a$ and $b$, $\sqrt{a+b} \neq \sqrt{a} + \sqrt{b}$?

Question

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?

Answer 1

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!

Answer 2

Supose that there is an equality and that $ab\neq 0$. Then square both sides. Then conclude when you obtain a contradiction.

Answer 3

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.

Answer 4

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 $b=0$.

Answer 5

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.

Answer 6

Yes you can!

If it's true then it should be true for $$a=b=1,$$ which gives $$\sqrt2=2,$$ which is a contradiction.

Answer 7

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 $a,b:$

$(a+b)^{1/2} \lt a^{1/2}+b^{1/2}.$