How to prove an inequality with a direct proof?

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SUMMARY

The discussion focuses on proving the inequality $$a^{n} - b^{n} \leq na^{n-1}(a-b)$$ for real numbers $$a$$ and $$b$$ where $$0 < b < a$$ and $$n$$ is a positive integer. A direct proof approach is suggested by factorizing the expression $$a^n - b^n$$ into $$(a-b)(a^{n-1} + \ldots + b^{n-1})$. This factorization allows for estimating the size of the second factor, which is crucial for completing the proof.

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  • Understanding of real numbers and inequalities
  • Familiarity with polynomial factorization techniques
  • Knowledge of mathematical induction for positive integers
  • Basic concepts of limits and continuity in calculus
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  • Study polynomial factorization methods, specifically for expressions like $$a^n - b^n$$
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  • Explore direct proof techniques in mathematical analysis
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Students in mathematics, particularly those studying real analysis or calculus, as well as educators looking for effective proof strategies in inequality problems.

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
 

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