Answer:
See explanation
Step-by-step explanation:
Let x, y ∈ R.
If x and y are both positive, then [tex]\sqrt{x+y}\neq \sqrt{x}+\sqrt{y}[/tex]
Suppose that
[tex]\sqrt{x+y}=\sqrt{x}+\sqrt{y}[/tex]
Square both sides of this equation:
[tex](\sqrt{x+y})^2=(\sqrt{x}+\sqrt{y})^2\\ \\x+y=(\sqrt{x})^2+2\sqrt{x}\sqrt{y}+(\sqrt{y})^2\\ \\x+y=x+2\sqrt{x}\sqrt{y}+y\\ \\2\sqrt{x}\sqrt{y}=0[/tex]
Then
[tex]\sqrt{x}=0\ \text{or}\ \sqrt{y}=0\\ \\x=0\ \text{or}\ y=0[/tex]
But x and y are both positive, so the assumption is false
