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.
 A: 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?
A: $$f(x)=x + \frac{x}{x^2 + 1} + \frac{x(x + 4)}{x^2 + 2} + \frac{2(x + 2)}{x(x^2 + 2)}$$
$$f(x)=\frac{x^6+x^5+8 x^4+3 x^3+12 x^2+2 x+4}{x \left(x^2+1\right) \left(x^2+2\right)}$$
$$f'(x)=\frac{ x^{10}+x^8-2 x^6-8 x^4-12 x^2-8}{\big[x \left(x^2+1\right) \left(x^2+2\right)\big]^2}$$
$$x^{10}+x^8-2 x^6-8 x^4-12 x^2-8=\left(x^3-2 x^2+2 x-2\right) \left(x^3+2 x^2+2 x+2\right) \left(x^4+x^2+2\right)$$

*

*The last factor does not show any real root.

*The first cubic  shows only one real root $x_1$

*The second cubic  shows only one real root $x_2$ which in fact $x_2=-x_1$
So, let us solve $x_1$ using the hyperbolic method
$$x_1=\frac{2}{3} \left(1+\sqrt{2} \sinh \left(\frac{1}{3} \sinh ^{-1}\left(\frac{17}{2
   \sqrt{2}}\right)\right)\right)$$
Rrplacing, we have the simple $f(x_1)=5$ and $f(x_2)=-3$.
A: 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 $2$
