I have been trying to find a lower bound for $x>1$, $c>0$:
$$ \Large(x^c-1)^{1/c} $$
My strategy is to find a lower bound for $(x^c-1)^{1/c}$ which can hopefully get rid of some of the $c$ powers. My guess is it should be greater than something along the lines of $x-\frac{1}{x^{c-1}}$.
If $c>1$ I can use convexity of $x^{1/c}$ I can write $(x-1)^{1/c}\ge x^{1/c}-\frac{1}{c}(x-1)^{1/c-1}$, however while it allows me cancel out one pair of $c$ and $\frac{1}{c}$ it just creates a new one.
Do you have any ideas what sort of inequalities I might benefit from having a look at?
I ended up solving this myself. There are two bounds, depending on $c$ being greater than, or less than 1:
$$(x^c-1)^\frac{1}{c} \ge \begin{cases} x\left(1-\frac{1}{cx^c}\right) & \text{if } 0 < c \le 1 \\ x\left(1-\frac{1}{c(x^c-1)}\right) & \text{if } 0 < c \end{cases}$$
This first bound follows from convexity, and the second one follows from the binomial theorem like this:
$$ \begin{align} (x^c-1)^\frac{1}{c} &= \sum_k{1/c\choose k}(-1)^k(x^c)^{1/c-k}\\ &=x-\frac{1}{c}x^{1-c}\left(1+\frac{1-1/c}{2!}x^{-c}+\frac{(1-1/c)(2-1/c)}{3!}x^{-2c}+\dots\right)\\ &\ge x-\frac{1}{c}x^{1-c}\left(1+x^{-c}+x^{-2c}+\dots\right)\\ &= x-\frac{1}{c}x^{1-c}\left(1+\frac{1}{x^c-1}\right)\\ &= x\left(1-\frac{1}{c(x^c-1)}\right) \end{align}$$
Experimentally, the bound is a lot better for large $c$'s than small ones. However since the second one is strictly smaller than the first one, it might be simpler to just always use that one.
If we underestimate the binomial sum more tightly as $x-\frac{1}{c}x^{1-c}\left(1+\frac{1}{2}x^{-c}+\frac{1}{3}x^{-2c}+\dots\right)$ we get the bound $\ge x\left(1+\frac{\log(1-x^{-c})}{c}\right)\ge x\left(1-\frac{1}{c(x^c-1)}\right)$ which is the same as before.
As a corollary this gives $$1+\frac{\log2}{c(\log_2x)^{c-1}}\le\frac{x}{2^{(\log_2x)^c-1)^{1/c}}} \le 1+\frac{\log2}{(c-\epsilon)(\log_2x)^{c-1}}$$ for $0<\epsilon<c\neq1$ and $x$ large enough, which was the problem I was interested in originally.