An inequality without studying variations How can one prove the following inequality using only concavity of $\log$ function: for any $\alpha>0$, there exists a constant $c_\alpha >0$ such that
$$\frac{\alpha x-1}{\log(\alpha x)} \le c_\alpha \frac{x-1}{\log x}, \qquad x>0.$$
We can prove it by studying the variation of a suitable function, but I found a claim that can be proved using only concavity of $\log$ by discussing the cases $\alpha>1$ and $\alpha<1$.
The point is to take out the constant $\alpha$ from the function on L.H.S and obtain just a multiplicative constant instead.
 A: The claim can be in fact sharpened to the statement that the least possible value of $c_a$ is:
$$
\max(1,a),
$$
since the function $\phi(x)=\frac{a x-1}{\log(a x)}\cdot\frac{\log x}{x-1}$ is monotonic and:
$$
\lim_{x\to0}\phi(x)=1;\quad
\lim_{x\to\infty}\phi(x)=a.
$$
For $a<1$ the resulting inequality 
$$
\frac{ax-1}{\log(ax)}\le \frac{x-1}{\log(x)}
$$
can be proved from concavity of the $\log(x)$ function. 
We start with the following inequality valid for arbitrary concave function $f(x)$:
$$
(x_3-x_2)f(x_1)+(x_2-x_1)f(x_3)\le(x_3-x_1)f(x_2),\tag1
$$
where $x_1< x_2< x_3$. 
Let now $f(x)=\log(x)$, $a<1$ and consider the following 3 cases:
I: $0<x<1$
Setting in (1) $x_1=ax,x_2=x,x_3=1$ one obtains
$$
(1-x)\log(ax)\le(1-ax)\log(x)\implies \frac{ax-1}{\log(ax)}\le\frac{x-1}{\log(x)}.
$$
II: $1<x<\frac1a$
Setting in (1) $x_1=ax,x_2=1,x_3=x$ one obtains
$$
(x-1)\log(ax)+(1-ax)\log(x)\le0\implies \frac{ax-1}{\log(ax)}\le\frac{x-1}{\log(x)}.
$$
III: $\frac1a<x$
Setting in (1) $x_1=1,x_2=ax,x_3=x$ one obtains
$$
(ax-1)\log(x)\le (x-1)\log(ax)\implies \frac{ax-1}{\log(ax)}\le\frac{x-1}{\log(x)}.
$$
Above we have used the fact that the product $\log(x)\log(ax)$ is positive in cases I and III and negative in the case II.
For $a>1$ the same argument applied to the convex function $x\log(x)$ results in the inequality:
$$
\frac{ax-1}{\log(ax)}\le a\frac{x-1}{\log(x)}.
$$
