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Let {$a_{n}$} be an arithmetic sequence with positive terms. Prove that for any $n \in \mathbb{N}$ $$\sqrt{a_{1}a_{n}}\leq\sqrt[n]{a_{1}a_{2}\cdot...\cdot a_{n}}$$ When proving that using induction, induction step would be if it is true for some n, we show that $$\sqrt{a_{1}a_{n}a_{n+1}}\leq\sqrt[n+1]{a_{1}a_{2}\cdot...\cdot a_{n}a_{n+1}}$$ or $$\sqrt{a_{1}a_{n+1}}\leq\sqrt[n+1]{a_{1}a_{2}\cdot...\cdot a_{n}a_{n+1}}$$ and why? Also solutions for this problem are appreciated.

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  • $\begingroup$ Only your last inequation is valid as an induction step. $\endgroup$
    – user65203
    Jan 5, 2021 at 8:12
  • $\begingroup$ @talbi: why $a_{n+2}$ ?? $\endgroup$
    – user65203
    Jan 5, 2021 at 8:14
  • $\begingroup$ @YvesDaoust Oops, you are correct - it's fine how it is. My bad! $\endgroup$
    – talbi
    Jan 5, 2021 at 8:14
  • $\begingroup$ @YvesDaoust I thought so but my question is why is that $\endgroup$
    – Vojtie
    Jan 5, 2021 at 8:20
  • $\begingroup$ By arithmetic sequence do you mean the terms are of the form $a_j=a+jd$ for some $a,d$? $\endgroup$ Jan 5, 2021 at 8:22

1 Answer 1

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First show that \begin{align} a_1 a_n \le a_k a_{n-k+1} \end{align} for any $k=1,\ldots,n$. By using the assumption that $\{a_n\}$ is an arithmetic sequence with positive terms, it can be easily shown. (Equality holds when $k=1$, $k=n$, or $d=0$, where $d$ is the common difference.)

Then multiplying the above inequalities for $k=1,\ldots,n$, we obtain \begin{align} (a_1 a_n)^n\le (a_1a_2\ldots a_n)^2. \end{align} (Note that all terms of $\{a_n\}$ appear twice. For example, $a_2$ appears when $k=2$ and $k=n-1$.) This completes the proof.

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