Saudi Arabian Mathematical Competitions

First solution. Assume that there exists $1 \leq i_{0}<n$ such that $a_{i_{0}}= a_{i_{0}+1}$. The sum of the two terms in the left hand side $a_{i_{0}-1} a_{i_{0}}\left(a_{i_{0}-1}-a_{i_{0}}\right)+ a_{i_{0}} a_{i_{0}+1}\left(a_{i_{0}}-a_{i_{0}+1}\right)$ is equal to $a_{i_{0}-1} a_{i_{0}+1}\left(a_{i_{0}-1}-a_{i_{0}-1}\right)$. This means that we can skip $i_{0}$ in the left hand side. This means that it is enough to prove this inequality when $a_{1}>a_{2}>\cdots>a_{n}>0$. In this case we have
$$\begin{aligned} & a_{1} a_{2}\left(a_{1}-a_{2}\right)+a_{2} a_{3}\left(a_{2}-a_{3}\right)+\cdots+a_{n-1} a_{n}\left(a_{n-1}-a_{n}\right) \\ & \quad=\frac{\left(a_{1}-a_{2}\right)^{2}}{\frac{1}{a_{1}}-\frac{1}{a_{2}}}+\frac{\left(a_{2}-a_{3}\right)^{2}}{\frac{1}{a_{2}}-\frac{1}{a_{3}}}+\cdots+\frac{\left(a_{n-1}-a_{n}\right)^{2}}{\frac{1}{a_{n-1}}-\frac{1}{a_{n}}} \\ & \quad \geq \frac{\left(\left(a_{1}-a_{2}\right)+\left(a_{2}-a_{3}\right)+\cdots+\left(a_{n-1}-a_{n}\right)\right)^{2}}{\left(\frac{1}{a_{1}}-\frac{1}{a_{2}}\right)+\left(\frac{1}{a_{2}}-\frac{1}{a_{3}}\right)+\cdots+\left(\frac{1}{a_{n-1}}-\frac{1}{a_{n}}\right)}=a_{1} a_{n}\left(a_{1}-a_{n}\right), \end{aligned}$$
by Cauchy-Schwarz inequality.
The equality holds here when
$$\frac{a_{1}-a_{2}}{\frac{1}{a_{1}}-\frac{1}{a_{2}}}=\frac{a_{2}-a_{3}}{\frac{1}{a_{2}}-\frac{1}{a_{3}}}=\cdots=\frac{a_{n-1}-a_{n}}{\frac{1}{a_{n-1}}-\frac{1}{a_{n}}},$$
which is equivalent to
$$a_{1} a_{2}=a_{2} a_{3}=\cdots=a_{n-1} a_{n} .$$
This implies that $n=2$, otherwise $a_{1}=a_{3}$ which is rejected. Therefore, in the general case, the equality holds when at least $n-2$ of the $n-1$ inequalities $a_{1} \geq a_{2} \geq \cdots \geq a_{n}$ are in fact equalities.

Second solution. We prove this inequality by induction on $n \geq 2$. For $n=2$, clearly both sides are the same. Assume this inequality true for $n$. Let $a_{1} \geq a_{2} \geq \ldots \geq a_{n} \geq a_{n+1}>0$. We have
$$\begin{aligned} & a_{1} a_{2}\left(a_{1}-a_{2}\right)+\cdots+a_{n-1} a_{n}\left(a_{n-1}-a_{n}\right)+a_{n} a_{n+1}\left(a_{n}-a_{n+1}\right) \\ & \quad \geq a_{1} a_{n}\left(a_{1}-a_{n}\right)+a_{n} a_{n+1}\left(a_{n}-a_{n+1}\right) \geq a_{1} a_{n+1}\left(a_{1}-a_{n+1}\right), \end{aligned}$$
since
$$\begin{aligned} a_{1} a_{n}\left(a_{1}-a_{n}\right) & +a_{n} a_{n+1}\left(a_{n}-a_{n+1}\right)-a_{1} a_{n+1}\left(a_{1}-a_{n+1}\right) \\ = & \left(a_{1}-a_{n}\right)\left(a_{n}-a_{n+1}\right)\left(a_{1}-a_{n+1}\right) \geq 0 \end{aligned}$$
The equality holds here when $a_{1}=a_{n}$ or when $a_{n}=a_{n+1}$ and there is equality in the hypothesis of induction. Therefore, in the general case, the equality holds when at least $n-2$ of the $n-1$ inequalities $a_{1} \geq a_{2} \geq \cdots \geq a_{n}$ are in fact equalities.