6
$\begingroup$

How to prove this inequality: $\dfrac{n(\ln(n)-2\ln(2))}{2\ln(n)(\ln(n)-\ln(2))^2} > 1$ for all $n\ge20$

I tried to apply this approach but I get a large first differentiate $u'(x)$ whose sign is not easy to determine

  • 0
    tried induction?2011-09-14
  • 0
    It is easy to see that it is true for sufficiently large numbers because the expression is the product of $n$ and a ratio of polynomials in $log(n)$. Since it is increasing, just calculate the smaller values.2011-09-14
  • 0
    @Arjang,No,I didn't2011-09-14
  • 2
    Arjang, I also suggested induction on pedja's earlier question http://math.stackexchange.com/questions/63283/proof-that-lnn2-lnn-1-n-for-all-n-in-mathbbn, but when mixedmath interrogated me on chat it became apparent that some calculus is required anyway. I wonder what is an elementary way to solve these types of problems with a particular base case (i.e. $n \ge 20$)?2011-09-14

1 Answers 1

5

Idea: The goal is to show that $f(x)$ is increasing. We could take the derivative, but in this case that will be tedious. Instead use logarithms. We have that $f(x)$ is positive and increasing if and only if $\log f(x)$ is increasing, so all we need to do is look at the logarithmic derivative and the value at $n=20$.

Proving the Inequality: Rewriting your above fraction, let $$g(n)=\frac{n\log\left(\frac{n}{4}\right)}{\log\left(n\right)\left(\log\left(\frac{n}{2}\right)\right)^{2}}.$$ then $$\log\left(g(n)\right)=\log n+\log\log\left(\frac{n}{4}\right)-\log2-\log\log n-2\log\log\left(\frac{n}{2}\right).$$ Taking the derivative we see that $$\frac{g^{'}(x)}{g(x)}=\frac{1}{x}+\frac{1}{x\log\left(\frac{x}{4}\right)}+\frac{-1}{x\log x}+\frac{-2}{x\log\left(\frac{x}{2}\right)}.$$ Now, notice that $\frac{1}{\log\left(\frac{x}{4}\right)}+\frac{-1}{\log x}>0$. Next, for $x\geq20$, $\log\left(\frac{x}{2}\right)\geq\log\left(10\right)>\log(e^{2})=2$ so that $1+\frac{-2}{\log\left(\frac{x}{2}\right)}>0.$ This implies that for $x\geq20$ $$\frac{g^{'}(x)}{g(x)}>0.$$ As $g(20)>1$ , the desired inequality follows.

  • 0
    ,I think that your definition of $g(n)$ is incorrect2011-09-14
  • 0
    @pedja: You are absolutely right! Thank you, there was a typo in that first line. I had switched the 2 and the 4! (Fortunately, for some reason that error didn't carry through the rest, and I reverted it in the second line. It is funny how things can end up when you type them out.)2011-09-14
  • 0
    ,Also it has to be proved that $\frac{x}{\ln(x)}$ is increasing for all $x\ge20$,but generally your idea is ok2011-09-14
  • 0
    @Pedja: Where in the proof do you think this is needed? I am inclined to disagree.2011-09-14
  • 0
    ,You have proved that fraction $\frac{g'(x)}{g(x)}$ is positive,but it can be positive and if $g'(x)$ and $g(x)$ were both negative2011-09-14
  • 0
    @pedja: $g(x)$ is clearly positive for $x> 4$ since everything in both the top and bottom of the fraction is positive. ($\log(x)$ is positive for $x>1$)2011-09-14
  • 0
    ,I can't accept that argument... "clearly" positive2011-09-14
  • 0
    ,That's sounds better...could you please correct definition of $g(n)$ and I will accept your answer2011-09-14