How to prove $1 - d^{-1/(d-1)} - d^{-d/(d-1)} > 0$ for $d \ge 2$? I want to show that
$$
f(d) := 1 - d^{-1/(d-1)} - d^{-d/(d-1)} > 0,
$$
for $d \ge 2$.
My solution is to show that $f(2)>0$,  $f(d) \to 0$ as $d \to \infty$, and that
$f(d)$ is decreasing on $[2, \infty)$ by taking derivative.
But I am wondering if there is a simpler proof? Maybe using some well-known inequality that I am not aware of?

You can use the generalisation of Bernoulli's inequality to non-integer exponents. For $p > 1$ and $x > 0$, we have
$$(1+x)^p > 1 + px.\tag{1}$$
Choosing $x = d-1$ and $p = \dfrac{d}{d-1}$ yields
$$d^{d/(d-1)} > 1 + d,\tag{2}$$
and multiplying $(2)$ with $d^{-d/(d-1)}$ yields the assertion (for $d > 1$).
