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I was trying to prove this inequality using induction, but couldn’t do.

Question: Suppose $a$ and $b$ are real numbers with $0 < b < a$. Prove that if $n$ is a positive integer, then:

$$a^n-b^n \leq na^{n-1}(a-b)$$

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You will want to use that $$a^n-b^n=(a-b)\sum_{k=0}^{n-1}a^{n-k-1}b^{k}$$

What can you say about the powers of $a,b$ given $0<b<a$?*

**SPOILER**

Since $0<b<a$, we have $0<b^k<a^k$

thus

$$\begin{align} a^n-b^n&=(a-b)\sum_{k=0}^{n-1}a^{n-k-1}b^{k}\\&<(a-b)\sum_{k=0}^{n-1}a^{n-k-1}a^{k}\\&=(a-b)\sum_{k=0}^{n-1}a^{n-1}\\&=(a-b)na^{n-1}\end{align}$$

Here is another somewhat related inequality:

$$ a^n – b^n > n(a-b)(ab)^{(n-1)/2} $$ where $ a > b > 0 $ and $ n \geq 1 $. Here is a simple proof:

$$a^n-b^n=(a-b)\sum_{k=0}^{n-1}a^{n-k-1}b^{k}$$ Now apply AM-GM inequality on summation terms:

\begin{align}

\sum_{k=0}^{n-1}a^{n-k-1}b^{k} &< n \sqrt[n]{\prod_{k=0}^{n-1}{a^{n-k-1}b^{k}}} \\

&= n \sqrt[n]{a^{\sum_{k=0}^{n-1}{(n-k-1)}} \, b^{\sum_{k=0}^{n-1}{k}}} \\

&= n \sqrt[n]{a^{n(n-1)/2} \, b^{n(n-1)/2}} \\

&= n (ab)^{(n-1)/2}

\end{align}

You may use the mean value theorem to show this:

Define $f(x) = x^n$ on $[b, a]$, clearly, $f(x)$ is differentiable in $(b, a)$ and continuous on $[b, a]$. By MVT, there exists $\xi \in (b, a)$ such that

$$a^n – b^n = f(a) – f(b) = f'(\xi)(a – b) = n\xi^{n – 1}(a – b) \leq na^{n – 1}(a – b)$$

as $\xi < a$.

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