The range of the function  $f\left(x\right)=\frac{x^2+x+2}{x^2+x+1},x\in R$, is  $(m,M]$ then  $m+M$ is

The given function is $f\left(x\right)=\frac{x^2+x+2}{x^2+x+1},x\in R$

Suppose that 

$\begin{array}{rcl}\frac{x^2+x+2}{x^2+x+1}& =& y\\ x^2+x+2& =& x^{2}y+xy+y=0\\ x^2\left(1-y\right)+x\left(1-y\right)+2-y& =& 0\end{array}$

This equation has real roots

its discriminant must be postive 

$\begin{array}{rcl}{b}^2-4ac& \ge & 0\\ {\left(1-y\right)}^2-4\left(1-y\right)\left(2-y\right)& \ge & 0\\ y^2-2y+1-4\left(2-y-2y+y^2\right)& \ge & 0\\ 3y^2-10y+7& \le & 0\end{array}$

It implies that $\begin{array}{rcl}3y^2-3y-7y+7& \le & 0\\ 3y\left(y-1\right)-7\left(y-1\right)& \le & 0\\ \left(y-1\right)\left(3y-7\right)& \le & 0\\ & & \end{array}$

Hence, $y\in \left(1,\frac{7}{3}\right)$