# An inequality concerning taking d-th roots of small real numbers

Let $1/2 < s< 1$ be a real number. How does one prove that $1-s \leq d(1-s^{1/d})$ for any positive integer $d$?

I can see the equality for $d=1$. I thought about showing that the derivative of $d(1-s^{1/d})$ with respect to $d$ is strictly positive, but didn't succeed. That is, I couldn't show that $d-ds^{1/d} +\log(s)s^{1/d} >0$ for every positive integer $d$.

I recall once seeing somewhere that $s^{1/d} \approx 1-s/d$ . If one makes this precise, one can prove the above inequality, I believe.

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Letting $u=1-s$, rewrite the inequality as $$1 - u \le \bigg[1 - \frac{u}{d}\bigg]^d,$$ which can be easily verified by taking $\log$ on both sides.
Elaborating. So we want to show $$\log (1 - u) \le d\log \bigg(1 - \frac{u}{d}\bigg).$$ Indeed, both sides tend to $0$ as $u \to 0^+$, so differentiating both sides of the inequality we want to show $$\frac{{ - 1}}{{1 - u}} \le d\frac{{ - 1/d}}{{1 - u/d}},$$ or $$\frac{1}{{1 - u/d}} \le \frac{1}{{1 - u}},$$ or $$u/d \leq u.$$ The original inequality is thus established.
By the way, the inequality is true for $0 < s < 1$ (hence above $0 < u < 1$). – Shai Covo Jun 20 '11 at 14:13