# Prove $\sqrt{a_{1}a_{n}}\leq\sqrt[n]{a_{1}a_{2}\cdot...\cdot a_{n}}$

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.

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

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.