Prove that $\forall x>0, \frac {x-1}{\ln(x)} \geq \sqrt{x} $.

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.

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 $e^t-e^{-t}=2\sum_{k=0}^{+\infty}\frac{t^{2k+1}}{(2k+1)!}\geqslant 2\frac{t^{2\cdot 0+1}}{(2\cdot 0+1)!}=2t$.