Inequality: $\dfrac{\ln x}{x^3-1}<\dfrac{x+1}{3(x^3+x)}$ for $x>0,\,x\ne 1$

[anemone]

Prove that $\dfrac{\ln x}{x^3-1}<\dfrac{x+1}{3(x^3+x)}$ for $x>0,\,x\ne 1$.

 

[anemone]

Solution of other:

Spoiler

For $x>0$, we define

$f(x)=\dfrac{(x^3-1)(x+1)}{(x^3+x)}-3\ln x=\dfrac{x^4+x^3-x-1}{(x^3+x)}-3\ln x$

Differentiating $f$ w.r.t. $x$, we have

$\begin{align*}f'(x)&=\dfrac{(4x^3+3x^2-1)(x^3+x)-(x^4+x^3-x-1)(3x^2+1)}{(x^3+x)^2}-\dfrac{3}{x}\\&=\dfrac{x^6+3x^4+4x^3+3x^2+1}{(x^3+x)^2}-\dfrac{3}{x}\\&=\dfrac{x^6-3x^5+3x^4-2x^3+3x^2-3x+1}{(x^3+x)^2}---(*)\end{align*}$

The polynomial $P(x)$ in the numerator on the right in (*) has the property that $P(x)=x^6P\left(\dfrac{1}{x}\right)$ and so $x^{-3}P(x)$ can be written as a cubic polynomial in $x+\dfrac{1}{x}$.

We have

$f'(x)=\dfrac{x^3}{(x^3+x)^2}\left(\left(x+\dfrac{1}{x}\right)^3-3\left(x+\dfrac{1}{x}\right)+4\right)$

as $x^3-3x+4=(x+1)(x-2)^2$, we obtain

$\begin{align*}f'(x)&=\dfrac{x^2+x+1}{(x^2+1)^2}\left(\left(x+\dfrac{1}{x}\right)-2\right)^2\\&=\dfrac{\left(\left(x+\dfrac{1}{x}\right)^2+\dfrac{3}{4}\right)(x-1)^2}{x^2(x^2+1)^2}\end{align*}$

so that $f'(x)>0$ for all $x>0$ while $f'(1)=0$.

Thus, $f(x)$ is a strictly increasing function of $x$ for all $x>0$. Hence in particular we have $f(x)>f(1)$ for $x>1$, and so

$\dfrac{\ln x}{x^3-1}<\dfrac{x+1}{3(x^3+x)}---(1)$ for $x>1$.

Replacing $x$ by $\dfrac{1}{x}$ in (1) we get

$\dfrac{\ln x}{x^3-1}<\dfrac{x+1}{3(x^3+x)}---(2)$ for $0<x<1$.

Inequalities (1) and (2) give the required result.