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

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!

• With the substitution $B = ( \frac{x}{2} +\sqrt{\frac{1}{27}+\frac{x^2}{4}} )^{1/3}$ (correspondingly, $x = B^3 - \frac{1}{27B^3}$ for $B > 0$), we have $f(x) = \frac{1}{3B} - B$ and $f'(x) = -\frac{3B^2}{1 - 3B^2 + 9B^4}$. Then it suffices to prove that $F(B) \ge 0$ for all $B > 0$ where $F$ is some polynomial. Commented Oct 4, 2020 at 4:52

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.
• @Albus Dumbledore Yes, of course. It's possible because $\frac{1}{27}+\frac{x^2}{4}>0.$ Commented Oct 4, 2020 at 15:06