So I want to prove that the strict concavity of the logarithm function implies $- \ln(x) \geq 1 - \ln(y) - (x/y)$, for positive x and y.
The definition I found of strict concavity is: $ \ln((1-\alpha)x + \alpha y) \geq (1-\alpha) \ln(x) + \alpha \ln(y)$, for $\alpha \in (0,1)$.
I tried just picking an $\alpha$, like $\frac{1}{e}$ but I just can't seem to get to the required form $- \ln(x) \geq 1 - \ln(y) - (x/y)$.
On
$-\ln(x) \geq 1 - \ln(y) - (x/y)$ can be rewritten as $ \ln (x/y) \le x/y -1 $ .
Now the feature of concavity which you state, $\ln((1-\alpha)x + \alpha y) \geq (1-\alpha) \ln(x) + \alpha \ln(y)$, defines the secant property of concave functions: pick two points $x$ and $y$ and all values on the secant $(1-\alpha) \ln(x) + \alpha \ln(y)$ are less than the respective values of the $\ln (.)$.
What is required here is another property of concave functions: at all points $x$ on the function where concavity holds, the function values for all $y$ are below the tangent values. I.e. $f(x) + f'(x) (y-x) \ge f(y)$.
Using that, write $x/y = z$ and establish $\ln z \le z-1 $. This can be done, using the stated concavity property, by using $f'(x) = \frac{1}{x}$ and writing $\ln z = \ln (1 + (z-1)) \le \ln 1 + \frac{1}{1}(z-1)) =z-1$.
This establishes the proof with concavity features only. $\qquad \square$
You don't need concavity.
Rearrange your equation to get
$1-x/y \le \ln y - \ln x = \ln(y/x) = -\ln(x/y) $ and this follows from $1-z \le -ln(z)$ or, replacing $z$ by $1-z$, $z \le -\ln(1-z)$ which is well-known.