MHB How to prove an inequality with a direct proof?

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To prove the inequality $$a^{n} - b^{n} \le na^{n-1}(a-b)$$ for real numbers a and b where $$0 < b < a$$ and n is a positive integer, start by factorizing the left side as $$(a-b)(a^{n-1} + \ldots + b^{n-1})$$. The next step involves estimating the size of the second factor, which consists of n terms, each less than or equal to a. This leads to the conclusion that the sum is less than or equal to na, allowing the inequality to hold. The approach provides a clear pathway to completing the proof directly.
Moodion
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Hello, I'm having trouble with an assigned problem, not really sure where to begin with it:

Prove that if $$a \in R$$ and $$b \in R$$ such that $$0 < b < a$$, then $${a}^{n} - {b}^{n} \le {na}^{n-1}(a-b)$$, where n is a positive integer, using a direct proof.

Pointers or the whole proof would be appreciated (might require some explanation afterwards!)

Thanks
 
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Moodion said:
Hello, I'm having trouble with an assigned problem, not really sure where to begin with it:

Prove that if $$a \in R$$ and $$b \in R$$ such that $$0 < b < a$$, then $${a}^{n} - {b}^{n} \le {na}^{n-1}(a-b)$$, where n is a positive integer, using a direct proof.

Pointers or the whole proof would be appreciated (might require some explanation afterwards!)

Thanks
Hi Moodion and welcome to MHB!

Try factorising $a^n-b^n$ as $(a-b)(a^{n-1} + \ldots + b^{n-1})$ and then estimate the size of the second factor.
 
Thanks for the hint, just what I needed
 
I was reading a Bachelor thesis on Peano Arithmetic (PA). PA has the following axioms (not including the induction schema): $$\begin{align} & (A1) ~~~~ \forall x \neg (x + 1 = 0) \nonumber \\ & (A2) ~~~~ \forall xy (x + 1 =y + 1 \to x = y) \nonumber \\ & (A3) ~~~~ \forall x (x + 0 = x) \nonumber \\ & (A4) ~~~~ \forall xy (x + (y +1) = (x + y ) + 1) \nonumber \\ & (A5) ~~~~ \forall x (x \cdot 0 = 0) \nonumber \\ & (A6) ~~~~ \forall xy (x \cdot (y + 1) = (x \cdot y) + x) \nonumber...

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