# Proving Inequality using Induction $a^n-b^n \leq na^{n-1}(a-b)$

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:

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

#### Solutions Collecting From Web of "Proving Inequality using Induction $a^n-b^n \leq na^{n-1}(a-b)$"

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 $\xi < a$.