Inequality of cauchy

Question

I found the following inequality in a book, dubbed "Cauchy's inequality for subunitary exponent":

$$ sb^{s-1}(b-a) < b^s - a^s < sa^{s-1}(b-a)\text{ } (0<a<b, s \in (0,1))$$

I tried proving it, but i didn't have much luck. I defined the function $$f(s) = b^s - a^s - sb^{s-1}(b-a).$$ in order to prove the first inequality. Obviously $f(0) = 0$, and so if I could prove that $f'(s) > 0$ when $s \in (0,1)$, I would be done. However, I didn't have much luck proving this inequality either.

How could one prove this?

Hardy proves it using basic algebra in his classic A Course of Pure Mathematics. The same proof is given in this answer : https://math.stackexchange.com/a/1782225/72031 — Paramanand Singh, Oct 17 '18 at 09:17

score 1 · Accepted Answer · answered Oct 15 '18 at 08:26

1

Hint: Using mean value theorem $$ \frac{b^s-a^s}{b-a} = sc^{s-1} $$ for $c\in(a, b)$.

Can you continue?

answered Oct 15 '18 at 08:26

Jakobian

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Inequality of cauchy

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