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Let $a>0$, $n,p \in \mathbb{N}$, $p<n$

$\sqrt[\leftroot{-1}\uproot{2}n]{a^p} \leq 1+\frac{p}{n} (a-1)$

I tried to do it by induction, with the first step for $a=2, n=2, p=1$. $\sqrt{2^1} \leq 1+ \frac{1}{2}(2-1)$. However I do not know how to go on with the induction. Would appreciate help

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  • $\begingroup$ hint: $a = (a-1) + 1$ $\endgroup$
    – across
    Commented Apr 9, 2021 at 2:49
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    $\begingroup$ Isnt this a modification of bernoullis inequality? $\endgroup$
    – Albus Dumbledore
    Commented Apr 9, 2021 at 2:50

1 Answer 1

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Apply the AM-GM inequality: $$ (x_1\cdot x_2\cdots x_n)^{\frac{1}{n}}\leq \frac{x_1+x_2+\ldots+x_n}{n} $$ to $\Big(\underbrace{a\cdots a}_{\text{p times}} \cdot \underbrace{1\cdots 1}_{\text{(n-p) times}} \Big)$ gives $$ a^{\frac{p}{n}}\leq \frac{pa+n-p}{n}=\frac{p}{n}a+\frac{n-p}{n}=1+\frac{p}{n}(a-1). $$

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  • $\begingroup$ thank you so much $\endgroup$
    – Inocenciaa
    Commented Apr 12, 2021 at 19:22

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