let $ \{a_n\} $ be an arithmetic sequence of positive numbers.
prove that for all $ n\ge2 $:
$ \frac{a_{1}+a_{n}}{2}\ge\sqrt[n]{a_{1}a_{2}a_{3}\dots a_{n}}\ge\sqrt{a_{1}a_{n}} $
I managed to solve the left part of the inequality by using the sum of the arithmetic sequence and the EM-GM inequality.
I'm having issues with the right side, tried using ln on everything and induction but the algebra was too tough. Also tried comparing with the harmonic mean Any other methods to solve this with calculus 1 tools(haven't learned about $ \Gamma $ function yet) ?
Let $a_n = a + (n-1)d$.
Then for $0 < m < n-1$ we have:
$$a_{1 + m}a_{n-m} = (a_1 + md)(a_n -md) \\= a_1a_n+(a_n-a_1)md-(md)^2 =a_1a_n +md((n-1-m)d)\ge a_1a_n$$
Now consider $(\sqrt[n]{a_1a_2\dots a_n})^2 = \sqrt[n]{a_1a_2\dots a_n}\sqrt[n]{a_na_{n-1}\dots a_1} = \sqrt[n]{(a_1a_n)(a_2a_{n-1})\dots(a_na_1)}$.
By our result above, that expression cannot be less than $\sqrt[n]{(a_1a_n)^n} = a_1a_n$.
Therefore $\sqrt[n]{a_1a_2\dots a_n} \ge \sqrt{a_1a_n}$.