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

Question

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:

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

Answer 1

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

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

Answer 2

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 $\xi < a$.

Answer 3

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}

Answer 4

The proof proceeds by induction.

(1) First base case $(n = 1)$: here $$a^1 −b^1 \le 1 \times a^0(a−b) = a − b$$, so the statement holds.

(2) Inductive Hypothesis: Now suppose that $a^k − b^k < ka^{k−1}(a − b)$, where $0< b < a$ for some $k \in \mathbb{N}$.

(3) Inductive Step: We want to show that

$$a^{k+1} − b^{k+1} < (k+1)a^{k}(a − b)$$

Then

$$\begin{align*} a^{k+1} − b^{k+1} &= a^{k+1}- a^k \times b + a^k \times b − b^{k+1} \\ &= a^k(a − b) + b(a^k − b^k) \\ &\leq a^k(a − b) + b \times ka^{k−1}(a− b) \quad\text{(By Inductive Hypothesis)}\\ &< a^k(a − b) + a \times ka^{k−1}(a − b) \quad\text{(Since a > b)}\\ &= a^k(a − b) + ka^{k}(a − b) \\ &= (k+1)a^k(a − b) \end{align*}$$

Hence by induction $a^n-b^n \leq na^{n-1}(a-b)$ is true $\forall n \in \mathbb{N}$.