How to say $\text {log}\ \ a^{-1} \geq 1-a$ from the concavity of $\text{log}(\cdot)$

Question

I am reading a paper and confront the following small trick:

$\text {log}\ \ a^{-1} \geq 1-a$, where $0\leq a \leq1$. By the concavity of $\text{log}(\cdot)$.

From the formula:

$f(\alpha x_1+(1-\alpha)x_2)\geq \alpha f(x_1)+(1-\alpha)f(x_2)$

I have no idea how to understand that trick.