For a positive integer $n\geq 2$ with divisors $1=d_1<d_2<\cdots<d_k=n$, prove that $d_1d_2+d_2d_3+\cdots+d_{k-1}d_k<n^2$

Question

IMO 2002 P4
Let $n\geq 2$ be a positive integer with divisors $1=d_1<d_2<\cdots<d_k=n$.
Prove that $d_1d_2+d_2d_3+\cdots+d_{k-1}d_k$ is always less than $n^2$, and determine when it is a divisor of $n^2$

I am trying this question but I run out of ideas, could someone give a little hint or a suggestion? Please, without giving me the solution.

I am trying to use the fact that the product of $d_i$*$d_{i+1}$ is a divisor of $n^2$ (and they are all different) and maybe try to use the formula for the sum of divisors to see if this specific sum is less than $n^2$

Answer 1

Hint 1: How large can $d_{k-1}$ be as a function of $n$? What about $d_{k-2}$?

Hint 2: Let $p$ be the smallest prime factor of $n$. What can you say about $d_{k-1}$ in terms of $n,p$? What’s the largest (proper) divisor of $n^2$?