lemma: $0 < x \leq y$ implies $\frac{y^2}x - \frac{x^2}y \geq 3(y-x)$. proof: rearranges to $(y-x)^3 \geq 0$. note that assuming $a_1 \leq \dots \leq a_n$ makes our job not easier so do this. then we claim $(a_1, \dots, a_n)$ and $(a_n, \dots, a_1)$ work. proof: take the differences as terms of the form $\frac{a_i^2}{a_{i-1}} - \frac{a_{i-1}^2}{a_i}$ and apply the lemma. (absolute value disappears because of our ordering)

According to your lemma it does not mean "at least"....you gave a lower bound....but the question asks for an upper bound right?...

Same as #2, except i proved the lemma using am-gm.