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
 
First trick I learned this one a long time ago and have used it to entertain and amuse young kids. Ask your friend to write down a three-digit number without showing it to you. Then ask him or her to rearrange the digits to form a new three-digit number. After that, write whichever is the larger number above the other number, and then subtract the smaller from the larger, making sure that you don't see any of the numbers. Then ask the young "victim" to tell you any two of the digits of the...

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