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How to prove, for every $x>0$ $$\dfrac{1}{\ln^2\left(1+\dfrac{1}{x}\right)} \dfrac{1}{(x+1)x}-1>0.$$

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Does $ln^2()$ mean square of the logarithm or take the logarithm twice? – Henry Feb 8 '12 at 8:45
Which operator is between first and second fraction ? – pedja Feb 8 '12 at 8:45
@pedja multiple – 89085731 Feb 8 '12 at 8:51
@Henry $(ln())^2$ – 89085731 Feb 8 '12 at 8:52
Define $f(x)$ as : $$f(x)=\frac{1}{x(x+1)}-(\ln(x+1)-\ln x)^2$$ then calculate $f'(x)$ and try to show that $f'(x)>0$ for all $x>0$ . – pedja Feb 8 '12 at 9:07

Here are some hints or ideas.

Exploring a few equivalent algebraic reformulations, for $x>0$, $$ \dfrac{1}{\ln^2\left(1+\dfrac1{x}\right)} \cdot \dfrac1{(x+1)x} - 1 > 0 $$ $$ \iff \ln^2\left(1+\dfrac1{x}\right) < \dfrac1{(x+1)x} = \dfrac1{x}-\dfrac1{x+1} $$ $$ \iff \ln\left(1+\dfrac1{x}\right) < \dfrac1{\sqrt{x(x+1)}} $$ $$ \iff \ln\left(1+u\right) < \dfrac{u}{\sqrt{u+1}} \quad \text{for} \quad u=\frac1x>0. $$ The last of these is fairly easy to work with. Setting $$ f(u)=\ln\left(1+u\right), \quad g(u)=\dfrac{u}{\sqrt{u+1}}=(1+u)^\frac12-(1+u)^{-\frac12} $$ we see that $f(u),g(u)\geq 0$ for $u>0$, that $f(0)=g(0)=0$, that $f(u),g(u)\approx u$ for $u\approx0$, and that $$ \lim_{u\rightarrow\infty}\frac{f(u)}{g(u)}=0 $$ (using L'Hôpital's rule) so graphing the two functions should get us almost there. For this, we probably only need the first one or two derivatives, $$ f'(u)=\left(1+u\right)^{-1}, \quad g(u) = \frac{ (1+u)^{-\frac12} - (1+u)^{-\frac32} }{2} = \frac12u(1+u)^{-\frac32} $$ which at zero are $$ f'(0)=g'(0)=1 \quad\text{and} $$ $$ f''(0)=-1,\quad g''(0)=-\frac12. $$ This should be enough to produce a convincing graph something like below,

enter image description here

made with sage (online).

G = plot(log(1+u),      (u,0,10),color='red')
G+= text('f(u)',           (3,1),color='red')
G+= plot(u*(1+u)^(-1/2),(u,0,10),color='blue')
G+= text('g(u)',           (1,1),color='blue')

For the full Taylor series (which we probably don't want), we would need the $n^\text{th}$ derivatives $$ f^{(n)}(u)=\frac{(-1)^{n-1}(n-1)!}{(1+u)^{n}} $$ and $$ g^{(n)}(u) =\frac12a_{n-1}(1+u)^{\frac12-n}-a_n(1+u)^{-\frac12-n} $$ for $n>0$, where $$ a_n =(-1)^n \left(\frac12\right) \left(\frac32\right) \cdots \left(\frac{2n-1}2\right) =(-1)^n \frac{(2n)!}{2^{2n}n!} =\left(n-\frac12\right)a_{n-1} $$ so that $$ g^{(n)}(u) =a_{n-1}\frac{\frac12(1+u)-\left(n-\frac12\right)}{(1+u)^{n+\frac12}} =a_{n-1}\frac{1+\frac{u}2-n}{(1+u)^{n+\frac12}}, $$ but there is probably an easier way than using the full Taylor series.

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Define the function: $$f(x)=\frac{1}{ln^2(1+\frac{1}{x})}\frac{1}{x(1+x)}$$ The reciprocal of this function must be less than $1$. Note that $$\lim_{x=\infty}{f(x)^{-1}}=1$$ and that $$\lim_{x=0}{f(x)^{-1}}=0$$ Because the derivative of the $f(x)^{-1}$ is never zero, this means that $1$ is an asymptote because there isn't a maximum between zero and $+\infty$

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