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On page 10 of Rudin's Principles of Mathematical Analysis, Rudin makes a claim that the identity

$b^{n} - a^{n} = (b - a)(b^{n-1} + b^{n-2}a + \cdots + a^{n-1})$

yields the inequality $b^{n} - a^{n} < (b - a)nb^{n-1}$ when $0 < a < b$, for real $a$ and $b$, and natural $n$.

Is this result obvious or does it actually require proving? Rudin gives no explanation for why it's true and I failed at finding any obvious reasons for it being true.

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    $\begingroup$ If $0<a<b$ then $b^{n-1}+b^{n-2}a+\cdots+a^{n-1}<b^{n-1}+b^{n-1}+\cdots+b^{n-1}=nb^{n-1}$ $\endgroup$
    – J. W. Tanner
    Commented May 31, 2020 at 3:55
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    $\begingroup$ Do you agree that $b^{n-i}a^{i-1}<b^{n-1}$ since $0<a<b$? $\endgroup$
    – user10575
    Commented May 31, 2020 at 3:56
  • $\begingroup$ Hint: if $0<a<b$, then $a^j<b^j$ for any positive integer $j$. Use this fact in the second pair of parentheses. $\endgroup$
    – user122916
    Commented May 31, 2020 at 3:56
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    $\begingroup$ You need $n\ge2$; for $n=1$ you have equality. $\endgroup$
    – Angina Seng
    Commented May 31, 2020 at 4:55

1 Answer 1

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Hint:

If $0<a<b$, then $a^j<b^j$ for any positive integer $j$,

so $b^{n-1-j}a^j<b^{n-1-j}b^j=b^{n-1}$,

so $b^{n-1}+b^{n-2}a+\cdots+a^{n-1}<b^{n-1}+b^{n-1}+\cdots+b^{n-1}=nb^{n-1}$.

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