# Solving inequality analytically

• May 4th 2012, 02:57 AM
trowa
Solving inequality analytically
Dear all,

I have been trying to analytically prove a statement (which should hold) for a couple of days now, but am no longer making any progress. Any insights would therefore be very much appreciated. The statement is the following:

Given that $g\left( x,k\right) =\frac{x^{k}}{1-\alpha }$, where $\alpha =\frac{Ln\left( x^{k+1}-1\right) }{x^{k+1}-2}$, and $x\geq 2$ and $k\geq 1$, show that $g\left( x,k+1\right) >g\left( x,k\right)$.

My progress

Spoiler:

This is equivalent to showing $g\left( x,k+1\right) -g\left( x,k\right) >0,$ or

$x^{k+1}\frac{1}{1-\frac{Ln\left( x^{k+2}-1\right) }{x^{k+2}-2}}-x^{k}\frac{1}{1-\frac{Ln\left( x^{k+1}-1\right) }{x^{k+1}-2}}>0$

$\Longleftrightarrow x^{k}\left( \frac{x}{1-\frac{Ln\left( x^{k+2}-1\right)}{x^{k+2}-2}}-\frac{1}{1-\frac{Ln\left( x^{k+1}-1\right) }{x^{k+1}-2}}\right) >0$

Since $x^{k}>0$, it suffices to show that $\left( \frac{x}{1-\frac{Ln\left( x^{k+2}-1\right)}{x^{k+2}-2}}>\frac{1}{1-\frac{Ln\left( x^{k+1}-1\right) }{x^{k+1}-2}}\right)$

$\Longleftrightarrow\left( \frac{1}{x}\right) \left( \frac{Ln\left(x^{k+2}-1\right) }{x^{k+2}-2}-1\right) >\frac{Ln\left( x^{k+1}-1\right)}{x^{k+1}-2}-1$

Now I take out the logarithms:

$\Longleftrightarrow Ln\left( \left( x^{k+2}-1\right) ^{\left( \frac{1}{x}\right) \left( \frac{1}{x^{k+2}-2}\right) }\right)-\frac{1}{x}>Ln\left( \left( x^{k+1}-1\right) ^{\left( \frac{1}{^{x^{k+1}-2}}\right) }\right)-1$

$\Longleftrightarrow\frac{x-1}{x}>Ln\left( \frac{\left( x^{k+1}-1\right)^{\left( \frac{1}{^{x^{k+1}-2}}\right) }}{\left( x^{k+2}-1\right)^{\left( \frac{1}{x}\right) \left( \frac{1}{x^{k+2}-2}\right) }}\right)$

$\Longleftrightarrow Exp\left( \frac{x-1}{x}\right) >\left( x^{k+1}-1\right) ^{\left( \frac{1}{{x^{k+1}-2}}\right) }\frac{1}{\left(x^{k+2}-1\right) ^{\left( \frac{1}{x}\right) \left( \frac{1}{x^{k+2}-2}\right) }}$

Now it suffices to show that:

$Exp\left( \frac{x-1}{x}\right) >\frac{3}{2} >\left( x^{k+1}-1\right) ^{\left( \frac{1}{{x^{k+1}-2}}\right) }\frac{1}{\left(x^{k+2}-1\right) ^{\left( \frac{1}{x}\right) \left( \frac{1}{x^{k+2}-2}\right) }}$

First part:

$Exp\left( \frac{x-1}{x}\right) >\frac{3}{2}$ if $x>1.67$

Second part (here's where I am stuck):

$\frac{3}{2} >\left( x^{k+1}-1\right) ^{\left( \frac{1}{{x^{k+1}-2}}\right) }\frac{1}{\left(x^{k+2}-1\right) ^{\left( \frac{1}{x}\right) \left( \frac{1}{x^{k+2}-2}\right) }}$