Prove $\frac{x+1}{x-1}\ln x \geq 2$ 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 $\forall x \in \Bbb R_+\setminus \lbrace 0; 1\rbrace$, $f(x) \geq 2 \quad \blacksquare$
 A: 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 $f(x)\leq0.$
A: 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}.
