How to find the range of $y=\frac{x^2+1}{x+1}$ without using derivative? The only thing I know with this equation is $y=\frac{x^2+1}{x+1}=x+1-\frac{2x}{x+1}$.
Maybe it can be solved by using inequality.

Out of the given answers, one simple way is to get the $x$ in terms of $y$ and solve it :)
$y=\frac{x^2+1}{x+1}$
$$\Rightarrow yx + y = x^2 +1 \Rightarrow -x^2 + yx +(y-1)=0 \Rightarrow x = \frac{-y \pm\sqrt{y^2 + 4(y-1)}}{-2} \Rightarrow \frac{-y \pm \sqrt{(y-(-2-\sqrt5))(y - (\sqrt5-2))}}{-2}$$
The Lowest Value so will be obtained will be when the value under square root = 0, then in that case
$$ = \frac{-\sqrt5+2}{-2}$$
Now plug that value in the expression for $x$ Gives the minimal value of $y$:
