Show this inequality $\frac{n}{a_1 - a_0} + \frac{n - 1}{a_2 - a_1} + \cdots + \frac{1}{a_n - a_{n-1}} \ge \sum_{k=1}^n \frac{k^2}{a_k}$

Question

For $a_1, \ldots , a_n \in \mathbb{R}, a_1 < a_2 < \cdots <a_n$ and $a_i \ne 0$, show that

$\dfrac{n}{a_1 - a_0} + \dfrac{n - 1}{a_2 - a_1} + \cdots + \dfrac{1}{a_n - a_{n-1}} \ge \sum_{k=1}^n \dfrac{k^2}{a_k}$

where $a_0 = 0$.

I tried mathematical induction but not able to solve (not able to simplify n = k +1) expression.

The inequality mentioned in the chapters are

Cauchy-Schwarz Inequality

Weierstrass's Inequality

Tchebychev's Inequality

I think we need to use Tchebychev's Inequality to prove this but I'm not able to solve this.

Answer 1

We proceed by induction on $n$. For $n = 1$ we have $$ \frac{1}{a_{1} - a_{0}} \geq \frac{1}{a_{1}}, $$ which is clearly true, as $a_0 = 0$. Suppose the inequality holds for $n$. Then we have \begin{align*} \sum_{k = 1}^{n + 1}\frac{n + 2 - k}{a_{k} - a_{k - 1}} &= \sum_{k = 1}^{n}\frac{n + 1 - k}{a_{k} - a_{k - 1}} + \sum_{k = 1}^{n + 1}\frac{1}{a_{k} - a_{k - 1}} \\ & \geq \sum_{k = 1}^{n} \frac{k^{2}}{a_{k}} + \sum_{k = 1}^{n + 1}\frac{1}{a_{k} - a_{k - 1}}, \end{align*} by the induction hypothesis. It is therefore sufficient to prove that $$ \sum_{k = 1}^{n + 1}\frac{1}{a_{k} - a_{k - 1}} \geq \frac{(n + 1)^{2}}{a_{n + 1}}. $$ This is a straightforward application of the AM-HM inequality to the numbers $a_1 - a_0, \ldots, a_n - a_{n - 1}$.