How can $a\geq 2b\log b$ and $b\geq 9$ imply $a/\log a\geq b$?

Question

The following slick argument is used in a paper and I don't know how to prove it:

If $a\geq 2b\log b$ and $b\geq 9$, then $$ \frac{a}{\log a}\geq b.\tag{*}$$

It seems that one needs a upper bound for $\log a$, which is the part I don't know.

Could anyone prove (*)?

Answer 1

Since $b \ge 9$ and $a \ge 2 b \log(b)$, $a \ge 18 \log(9) > 18$.

If $a \ge 2 b \log(b)$ and $b > a/\log(a)$, we'd have $b \log(a) > a \ge 2 b \log(b)$, so $\log(a) > 2 \log(b)$, and thus $a > b^2$. But $ a/\log(a) > a^{1/2}$ for $a > 3$, so now

$$b > \frac{a}{\log(a)} > a^{1/2} > b$$

Contradiction!