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I'm trying to get back to maths after quite a pause, and I've solved a question in a way that does not satisfy me. I'm pretty sure there is a nicer way to go and I'd like your opinion.

$\forall x \in \Bbb R_+\setminus \lbrace 0; 1\rbrace$,$\quad $ $\frac{x+1}{x-1}\ln x \geq 2$

Let $f (x) = \frac{x+1}{x-1}\ln x$ defined on $\Bbb R_+\setminus \lbrace 0; 1\rbrace$

Then $$f'(x) = \frac{x^2-2x\ln x-1}{x(x-1)^2} = \frac{x-2\ln x-\frac 1x}{(x-1)^2}$$ which has the same sign as $$g(x) = x-2\ln x - \frac 1x$$

Since $$g'(x) = 1-\frac 2x + \frac{1}{x^2} = (1-\frac 1x)^2$$

$g$ is increasing over $\Bbb R_+$, and since $g(1)= 0$, $g(x)\leq 0$ for $0\leq x\leq 1$ and $g(x) \geq 0$ for $x\geq 1$ ; and so is $f'$.

Thus $f$ is decreasing on $]0; 1[$ and increasing on $]1; +\infty[$.

Now $$\lim_{x\to 1} \frac{\ln x}{x-1} =\lim_{x\to 1}\frac{\ln x -\ln 1}{x-1} = (\ln x)'(1) = 1 $$ and thus $$\lim_{x\to 1} f(x) = \lim_{x\to 1}(x+1)\frac{\ln x}{x-1}= 2$$

And thus $\forall x \in \Bbb R_+\setminus \lbrace 0; 1\rbrace$, $f(x) \geq 2 \quad \blacksquare$

krirkrirk
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2 Answers2

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Let $x>1$ and $f(x)=\ln{x}-\frac{2(x-1)}{x+1}.$

We need to prove that $f(x)\geq0$.

We see that $$f'(x)=\frac{(x-1)^2}{x(x+1)^2}>0,$$ which says $$f(x)>f(1)=0$$ For $0<x<1$ the proof is the same:

There, we need to prove that $f(x)\leq0.$

Michael Rozenberg
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  • That's way nicer indeed. How did you know you should study $\ln x - \frac{2(x-1)}{x+1}$ ? Did you see $f'(x)$ right away ? – krirkrirk Nov 10 '18 at 17:27
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    @krirkrirk I did not see it before, but it turned out good. I took this function because the derivative kills the $\ln$. – Michael Rozenberg Nov 10 '18 at 18:36
  • @krirkrirk This is a standard method; whenever you are trying to prove that $f(x) \ge g(x)$, one option is to try to prove that $h(x) = f(x)-g(x) \ge 0$ using the derivative. – Ovi Nov 13 '18 at 23:26
  • @Ovi of course, but it wasn't obvious for me what was the best choice for $f(x)$ and $g(x)$. – krirkrirk Nov 13 '18 at 23:32
  • @krirkrirk Ah sorry I didn't look too carefully; I'm sure it's obvious to you now, but what Michael did is the first thing to try since if you try to apply it directly to $\dfrac{x+1}{x-1}\ln x - 2$, the derivative would get extremely messy and the easiest way to make it simpler is what Michael did. – Ovi Nov 13 '18 at 23:35
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Define $$ f(x) = \frac{x+1}{x-1}\cdot\ln x \quad (x > 0, \ x \ne 1). $$ Then $f(1/x) = f(x)$, so in bounding $f$ we may assume that $x > 1$.

We prove $\fbox{f(x) > 2}$ for $x > 1$, by writing $x = e^u$, and proving \begin{equation} \tag{1}\label{ineq:1} \frac{e^u + 1}{e^u - 1} > \frac{2}{u} \quad (u > 0). \end{equation}

By simple rearrangement, \eqref{ineq:1} is equivalent to \begin{equation} \tag{2}\label{ineq:2} (2 - u)e^u < 2 + u \quad (u > 0). \end{equation} \eqref{ineq:2} is proved in several different ways here.

\eqref{ineq:1} and \eqref{ineq:2} are trivial for $u \geqslant 2$. For $u < 2$, they are equivalent to \begin{equation} \tag{3}\label{ineq:3} e^u < \frac{2 + u}{2 - u} \quad (0 < u < 2). \end{equation}

In an answer here, I argue that, because the graph of $1/t$ lies above the tangent at $t = 1$, $$ u < \int_{1-\frac{u}{2}}^{1+\frac{u}{2}}\frac{dt}{t} = \ln\left(1+\frac{u}{2}\right)-\ln\left(1-\frac{u}{2}\right) = \ln\frac{2+u}{2-u}, $$ whence \eqref{ineq:3}.

Calum Gilhooley
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