I am not able to solve this problem

if $\displaystyle a^2+b^2=1$ then find the range of $\displaystyle a+b$

what I am doing is

$\displaystyle (a+b)^2>0$

$\displaystyle a^2+b^2+2ab>0$ where $\displaystyle a^2+b^2=1$

$\displaystyle 1+2ab>0$

$\displaystyle 2ab>-1$

Case 2

$\displaystyle (a-b)^2>0$

$\displaystyle a^2+b^2-2ab>0$ where $\displaystyle a^2+b^2=1$

$\displaystyle 1-2ab>0$

$\displaystyle 2ab<1$

$\displaystyle -1<2ab<1$

$\displaystyle a+b=\sqrt{a^2+b^2+2ab}$

$\displaystyle a+b=\sqrt{1+2ab}$

$\displaystyle a+b=\sqrt{1-1}$ $\displaystyle -1<2ab<1$ to find minimum

$\displaystyle a+b=0$

$\displaystyle a+b=\sqrt{1+1}$ $\displaystyle -1<2ab<1$ to find maximum

$\displaystyle a+b=\sqrt{2}$

The range I am getting is $\displaystyle (0,\sqrt{2})$ but the answer is ($\displaystyle (-\sqrt{2},\sqrt{2})$