Infinite seq. of reals, for every n $ \in \mathbb{N}$ :$(a_{n-1}+a_{n+1})/2\geq a_n$. Prove $ \frac{a_0+a_{n+1}}{2}\geq \frac{a_1+a_2+...+a_n}{n} $

Question

I realized that in the case when $$\frac{a_{n-1}+a_{n+1}}{2} = a_n,$$ the seqence reduces to a sequence of natural numbers and so the inequality is trivially true: $$ \frac{a_{n+1}}{2}=\frac{a_1+a_2+...+a_n}{n} $$ I haven't been able to generalize the proof to cases when $$\frac{a_{n-1}+a_{n+1}}{2} > a_n$$Any help would be appreciated.

Answer 1

We can use induction here.

For $n=1$.

We need to prove that: $$\frac{a_2+a_0}{2}\geq a_1,$$ which is true by the given.

Now, by the assumption of the induction $$2(a_1+a_2+...+a_{n+1})\leq n(a_{n+1}+a_0)+2a_{n+1}$$ and it's enough to prove that: $$n(a_{n+1}+a_0)+2a_{n+1}\leq(n+1)(a_{n+2}+a_0)$$ or $$(n+1)a_{n+2}+a_0\geq(n+2)a_{n+1}$$ or $$(n+1)(a_{n+2}-a_{n+1})\geq a_{n+1}-a_0$$ or $$(n+1)(a_{n+2}-a_{n+1})\geq a_{n+1}-a_{n}+a_{n}-a_{n-1}+...+a_1-a_0,$$ which is true because $$a_{n+2}-a_{n+1}\geq a_{n+1}-a_n\geq a_{n}-a_{n-1}\geq...\geq a_1-a_0.$$