<s>[INST] 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.
 [/INST] Does not exist. 
Try $x\rightarrow-1^-$.
For $x>-1$ by AM-GM  we obtain:
$$\frac{x^2+1}{x+1}=\frac{x^2-1+2}{x+1}=x-1+\frac{2}{x+1}=$$
$$=x+1+\frac{2}{x+1}-2\geq2\sqrt{(x+1)\cdot\frac{2}{x+1}}-2=2\sqrt2-2.$$
The equality occurs for $x+1=\frac{2}{x+1},$ which says that we got a minimal value and the local minimal value.
Thus, the range for $x>0$ it's $[2\sqrt2-2,+\infty).$
For $x<-1$ we can get the range by the similar way. 
</s>