Tricky Algebra Question II

$\begin{aligned} f\left(x\right) &= x+\frac{x}{x^2+1}+\frac{\ x\left(x+4\right)\ }{x^2+2}+\frac{\ 2\left(x+2\right)\ }{\ x\left(x^2+2\right)} \\ &= \frac{x^3+x+x}{x^2+1}+\frac{\ x^3+4x^2+2x+4\ }{x\left(x^2+2\right)} \\ &= \frac{x\left(x^2+2\right)}{x^2+1}+\frac{4x^2+4\ }{x\left(x^2+2\right)}+\frac{\ x^3+2x\ }{x\left(x^2+2\right)} \\ &= \frac{x\left(x^2+2\right)}{x^2+1}+\frac{\ 4\left(x^2+1\right)\ }{x\left(x^2+2\right)}+1 \\ \end{aligned}$

By using Arithmetic-Geometric Mean Inequality ,

when $x>0$ , $f(x)\geq2\sqrt{\left(\frac{x\left(x^2+2\right)}{x^2+1}\right)\left(\frac{\ 4\left(x^2+1\right)\ }{x\left(x^2+2\right)}\right)\ }+1=5$ when $x<0$ , $f(x)\le-2\sqrt{\left(\frac{x\left(x^2+2\right)}{x^2+1}\right)\left(\frac{\ 4\left(x^2+1\right)\ }{x\left(x^2+2\right)}\right)\ }+1=-3$

From the above, we can found that $M+m$ is equal to $\boxed 2$ .