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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.

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    $\begingroup$ Have you tried induction $\endgroup$ – Jakobian Mar 21 at 16:58
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    $\begingroup$ What have you tried? You'll get a better response on this site if you tell us what you've tried and where you've gotten stuck. $\endgroup$ – jgon Mar 21 at 17:05
  • $\begingroup$ For $n=1$, $a_0 = -1$, $a_1 = 1$, your inequality is $\frac{1}{2} \geq 1$. It is wrong. Moreover, if one the $a_k$ is $0$, you can't even define the sum on the right. Please precise what numbers you consider. $\endgroup$ – TheSilverDoe Mar 21 at 17:10
  • $\begingroup$ Why $a_0$ can't be found in RHS? $\endgroup$ – Mostafa Ayaz Mar 21 at 21:51
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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 $a_1 - a_0, \ldots, a_n - a_{n - 1}$.

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