Proving $ 1+2f'(x)+\frac{2}{x(1+x^2)}\left(\frac{3x}{2}+f(x) \right)\ge \frac{6x^2}{1+8x^2} $.

Question

Put \begin{align*} f(x)=\left( -\frac{x}{2} +\sqrt{\frac{1}{27}+\frac{x^2}{4}} \right)^{1/3}-\left( \frac{x}{2} +\sqrt{\frac{1}{27}+\frac{x^2}{4}} \right)^{1/3} \end{align*} Prove that $$ g(x):=1+2f'(x)+\frac{2}{x(1+x^2)}\left(\frac{3x}{2}+f(x) \right)\ge \frac{6x^2}{1+8x^2} $$

My attempt

I put \begin{align*} A=\left( -\frac{x}{2} +\sqrt{\frac{1}{27}+\frac{x^2}{4}} \right)^{1/3}\quad B=\left( \frac{x}{2} +\sqrt{\frac{1}{27}+\frac{x^2}{4}} \right)^{1/3} \end{align*} and then \begin{align*} f'(x)=-\frac{1}{3}\frac{1}{A^2-AB+B^2} \end{align*} where $AB=\frac13$. But I don't know how to continue. I know that $g(x)$ is an even function. Via mathematica I find that $$\left[(1+x^2)g(x)\right]'\ge 0\quad \forall \,x>0$$ But I also can't prove this. Any hints? Thanks in advance!

Answer 1

The hint.

Let $f(x)=y$.

Thus, $$y^3+y+x=0,$$ which gives $$3y^2y'+y'+1=0$$ or $$y'=-\frac{1}{1+3y^2}$$ and we need to prove a polynomial inequality of one variable $y$.

I got that finally we need to prove that: $$y^2(6y^{14}+16y^{12}-10y^{10}+y^8+94y^6+94y^4+26y^2+1)\geq0,$$ which is obvious.