Find the minimum value of $f(x) = x + \frac{x}{x^2 + 1} + \frac{x(x + 4)}{x^2 + 2} + \frac{2(x + 2)}{x(x^2 + 2)}$ Find the minimum value of
$$f(x) = x + \frac{x}{x^2 + 1} + \frac{x(x + 4)}{x^2 + 2} + \frac{2(x + 2)}{x(x^2 + 2)}$$ for $x > 0.$
I'm not sure how to start this problem. I think AM-GM is the best way, but I'm not sure.

Here is your solution by A.M-G.M
$$f(x)=x+\frac{x}{x^2+1}+\frac{x(x+4)}{x^2+2}+\frac{2(x+2)}{x(x^2+2)}$$
$$f(x)=x+\frac{x}{x^2+1}+\frac{x^2+4x}{x^2+2}+\frac{2}{x^2+2}+\frac{4}{x(x^2+2)}$$
$$f(x)=\frac{x(x^2+2)}{x^2+1}+\frac{x^2}{x^2+2}+\frac{4x}{x^2+2}+\frac{2}{x^2+2}+\frac{4}{x(x^2+2)}$$
$$f(x)=\frac{x(x^2+2)}{x^2+1}+1+\frac{4x}{x^2+2}+\frac{4}{x(x^2+2)}$$
$$f(x)=1+\frac{x(x^2+2)}{x^2+1}+\frac{4(x^2+1)}{x(x^2+2)}\geq 1+2\sqrt{\frac{x(x^2+2)}{x^2+1}.\frac{4(x^2+1)}{x(x^2+2)}}$$
$$f(x)\geq 5$$
Equality occurs when$$\frac{x(x^2+2)}{x^2+1}=\frac{4(x^2+1)}{x(x^2+2)}$$
or$$x^3-2x^2+2x-2=0$$ which has a real root lying between $1$ and $2$
