Prove that $\forall x>0, \frac {x-1}{\ln(x)} \geq \sqrt{x} $. This inequality arose in this question  Prove that : $|f(b)-f(a)|\geqslant (b-a) \sqrt{f'(a) f'(b)}$ with $(a,b) \in \mathbb{R}^{2}$ :
$$\forall x>0, \frac {x-1}{\ln(x)} \geq \sqrt{x} $$
Can anybody find a way to prove it without calculus? Like using only inequalities like AM-GM,  Jensen or Cauchy - Schwarz? 
CLARIFICATION
I want a proof that does not involve setting $g(x)=\frac {x-1}{\ln(x)}- \sqrt{x}$, studying the derivative of $g$ and proving it is $\geq 0$ with this method.
The upvoted answers comply with this criterion.

Define $t:=\log x$. We have to show that for each $t\neq 0$, 
$$\frac{e^t-1}t\geqslant e^{t/2},$$
and, after having multiplied on both sides by $e^{-t/2}$, we are reduced to show that for each positive $t$, 
$$e^t-e^{-t}\geqslant 2t.$$
This inequality is easier to handle. We indeed recognize classical hyperbolic sine. 
If we know the power series expansion of exponential function, namely, $\exp(t)=\sum_{j=0}^{+\infty}\frac{t^j}{j!}$, the result follows because $e^t-e^{-t}=2\sum_{k=0}^{+\infty}\frac{t^{2k+1}}{(2k+1)!}\geqslant 2\frac{t^{2\cdot 0+1}}{(2\cdot 0+1)!}=2t$.
