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.

We have
$$f'(x) = \frac{(x^4+x^2+2)(x^3+2x^2+2x+2)(x^3-2x^2+2x-2)}{x^2(x^2+1)^2(x^2+2)^2}.$$
Because $x>0,$ so
$$f'(x) = 0 \Rightarrow  x = x_0 = \frac{\sqrt[3]{17+3\sqrt{33}}}{3}-\frac{2}{3\sqrt[3]{17+3\sqrt{33}}}+\frac{2}{3} \approx 1.5437$$
Therefore
$$f(x) \geqslant f(x_0) = 5.$$
So $f(x)_{\min} = 5,$ equality occur when $x=x_0.$

Note. The SOS form
$$f(x) = 5 + \frac{(x^3-2x^2+2x-2)^2}{x(x^2+1)(x^2+2)}  \geqslant 5.$$
P/s. How do center an image?
