为何替换n=1/n'无法推导不等式b-1≥n(b^(1/n)-1)

Let's break this down clearly, because it all boils down to respecting the conditions of the original inequality:

First, Recap the Original Inequality's Rules #

The given inequality:

$b^n - 1 \geq n(b - 1)$

has two non-negotiable constraints:

• $n$ is an integer (you specified this upfront)
• $b > 1$

This inequality holds reliably for all integers $n \geq 1$ and $b > 1$ (we’re taking it as given here, but you could prove it via induction or calculus if needed).

Why Your First Substitution Went Wrong #

You tried replacing $n$ with $\frac{1}{n'}$ (I’m assuming $n'$ is an integer, since the original $n$ was integer). Here’s the critical mistake:

• The original inequality only applies when the exponent $n$ is an integer. $\frac{1}{n'}$ is a fraction (between 0 and 1 for integer $n' > 1$), which violates the inequality’s core condition.

To make it worse, for real numbers $k$ where $0 < k < 1$, the inequality direction actually reverses! For $b > 1$ and $0 < k < 1$, we have:
$b^k - 1 \leq k(b - 1)$

That’s exactly why you got the opposite conclusion—you tried to apply an integer-only inequality to a non-integer exponent, where the relationship flips entirely.

Why the Correct Substitution Works #

The right approach is to substitute $b = t^n$ (where $t > 1$, since $b > 1$) into the original inequality. Let’s walk through it step by step:

1. Start with the valid original inequality (integer $n \geq 1$, $b > 1$):
   $b^n - 1 \geq n(b - 1)$
2. Let $t = b^{1/n}$ (so $b = t^n$, and since $b > 1$, $t > 1$ too). This substitution only changes the $b$ variable—$n$ remains an integer, which keeps the original inequality valid.
3. Plug $t = b^{1/n}$ into the original inequality (using $t$ as the new "b" value):
   $t^n - 1 \geq n(t - 1)$
4. Substitute back $t = b^{1/n}$:
   $(b^{1/n})n - 1 \geq n(b^{1/n} - 1)$
5. Simplify the left side, and you get exactly your target conclusion:
   $b - 1 \geq n(b^{1/n} - 1)$

This works because:

• We never messed with the integer status of $n$—it’s still the same integer from the original inequality.
• We only replaced $b$ with another value greater than 1 ($t^n$), which fits all the original inequality’s requirements for $b$.

Key Takeaway #

When working with inequalities that have restricted variables (like "n must be integer"), you can’t arbitrarily replace the restricted variable with a non-restricted value. But you can substitute other variables (like $b$ here) as long as you maintain all the original constraints (keeping $b > 1$, keeping $n$ integer).

内容的提问来源于stack exchange，提问作者mathnoob123