Inequality with $a_1,a_2, \cdots , a_n$

Let $$n \geqslant 2$$ be an integer and let $$1 be $$n$$ real numbers such that $$a_1+a_2+\dots+a_n=2n$$. Prove that$$a_1a_2\dots a_{n-1}+a_1a_2\dots a_{n-2}+\dots+a_1a_2+a_1+2 \leqslant a_1a_2\dots a_n.$$

I feel something is very wrong with my solution, but I can't figure it out. We will proceed with induction. We shall show the base case of $$2$$ [which I'm not including as it is not of any concern]. Assume the statement is true for $$n$$, hence we need to show it's true for $$n+1$$, that is $$2+a_1+a_1 a_2+...+a_1 a_2...a_n \leq a_1 a_2 ...a_n a_{n+1}$$ Note that the LHS is less than or equal to $$2a_1 a_2 ... a_n$$ using induction hypothesis [ that is $$a_1a_2\dots a_{n-1}+a_1a_2\dots a_{n-2}+\dots+a_1a_2+a_1+2 \leqslant a_1a_2\dots a_n$$]. Now if we can show $$2a_1 a_2 ... a_n \leq a_1 a_2 ...a_n a_{n+1}$$ Which is equivalent to showing $$2 \leq a_{n+1}$$ Now notice, $$a_1+a_2+\dots+a_n+ a_{n+1}=2n+2$$ and $$a_1+a_2+\dots+a_n=2n$$ From these two equation, we can get $$a_{n+1}=2$$. So $$2 \leq a_{n+1}$$ is true and hence we are done. Where is this proof wrong?

• See here: artofproblemsolving.com/community/c6h1989009 I hope it will help. Commented May 1, 2023 at 7:00
• @MichaelRozenberg I have gone through that thread, and all the (inductive) solutions there are different and much complex than mine. That's why I suspect that my solution is wrong, but still I can't pinpoint where it is wrong. Commented May 1, 2023 at 7:04

Writing the induction hypothesis clearly would mean: given any sequence $$a_1,...,a_n$$ Such that:

$$1 < a_1 \leq \ldots \leq a_n$$

$$a_1+...+a_n = 2n$$

Then $$2+a_1+\ldots+a_{n-1} \ldots a_1 \leq a_n\ldots a_1$$

Now if you take a sequence $$a_1,\ldots,a_{n+1}$$ satisfying the two conditions for $$n+1$$, then you don't necessarily have that $$a_1,\ldots,a_n$$ satisfies the conditions for $$n$$. Specifically you don't control $$a_1+...+a_n$$

The structure of a proof by induction is, you have a statement $$P(n)$$, in this case $$P(n)$$ is that all increasing n-tuples of sum $$2n$$ of integers strictly greater than 1 satisfy the given inequality. You prove the base case, and that for all n bigger than 2, you suppose that $$P(n)$$ is true and you show $$P(n+1)$$.

So if you take any increasing (n+1)-tuple of sum $$2n+2$$ of integers strictly greater than 1, you have no idea what the first n values have as sum, since that is just the hypothesis (not the conclusion) of the induction statement...

Hence you can't use the induction hypothesis on them directly as you did.

• "then you don't necessarily have that $a_1,…,a_n$ satisfies the conditions for n" why is that the case? How does $a_1,...,a_n$ lose their property one we extend it to $n+1$ terms? Commented May 1, 2023 at 7:18
• Why would $a_1+\ldots+a_{n+1}=2n+2$ imply that $a_1+\ldots+a_n=2n$ ? Commented May 1, 2023 at 8:06
• I hope my edit clarifies things Commented May 1, 2023 at 8:31