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.

For $x>1$, i.e. $\log x>0$, we can write $x=e^z$, and $z>0$. The inequality becomes
$$
z \leq e^{\frac{z}{2}}-e^{-\frac{z}{2}}.
$$
This is trivially true, for example by using the definition $$e^u = \sum_{k=0}^\infty \frac{u^k}{k!}.$$
The case $0<x<1$ can be treated in a similar fashion.
I know that this proof is not calculus-free, but it stems from the definition of the exponential function. This makes me think that it is not a convexity inequality like all those you refer to in your question. I might be wrong, of course.
