Showing that $a-ab^{-1/a}-b^{-1/a}\log(b)>0$ for $a>0$ and $b>1$?

Question

How can I show that $$a-ab^{-1/a}-b^{-1/a}\log(b)>0$$

given $a,b \in \mathbb{R}$ such that $a > 0$ and $b>1$?

Context: Finding the criteria for stability of a non-trivial steady-state, using the Jury Criterion, in a modified version of the Nicholson-Bailey model.

Thanks for any help.

Answer 1

$$a-ab^{-1/a}-b^{-1/a}\log(b)>0 \Leftarrow a-b^{-1/a}(a+\log(b))>0 \Leftarrow ab^{1/a}>a+\log(b)$$ This implies that $b^{1/a}>1+\log(b^{1/a})$. Let $x=b^{1/a}$. Then the inequality $x>1+\log(x)$ is true for all $x \in \mathbb{R}$ except when $x=1$. Since $a>0$ and $b>1$, $x>1$, so the result follows.