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

anemone
Gold Member
MHB
POTW Director
Messages
3,851
Reaction score
115
Prove that $\dfrac{\ln x}{x^3-1}<\dfrac{x+1}{3(x^3+x)}$ for $x>0,\,x\ne 1$.
 
Mathematics news on Phys.org
Solution of other:

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.
 
Suppose ,instead of the usual x,y coordinate system with an I basis vector along the x -axis and a corresponding j basis vector along the y-axis we instead have a different pair of basis vectors ,call them e and f along their respective axes. I have seen that this is an important subject in maths My question is what physical applications does such a model apply to? I am asking here because I have devoted quite a lot of time in the past to understanding convectors and the dual...
Thread 'Imaginary Pythagorus'
I posted this in the Lame Math thread, but it's got me thinking. Is there any validity to this? Or is it really just a mathematical trick? Naively, I see that i2 + plus 12 does equal zero2. But does this have a meaning? I know one can treat the imaginary number line as just another axis like the reals, but does that mean this does represent a triangle in the complex plane with a hypotenuse of length zero? Ibix offered a rendering of the diagram using what I assume is matrix* notation...

Similar threads

Back
Top