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

For $$a_1, \ldots , a_n \in \mathbb{R}, a_1 < a_2 < \cdots and $$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.

• Have you tried induction Mar 21, 2019 at 16:58
• 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.
– jgon
Mar 21, 2019 at 17:05
• 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. Mar 21, 2019 at 17:10
• Why $a_0$ can't be found in RHS? Mar 21, 2019 at 21:51
• Note that we likely need $a_1 > a_0 = 0$, and not just $a_i \neq 0$. Jan 5 at 15:01

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 $$a_1 - a_0, \ldots, a_n - a_{n - 1}$$.
• You do need $a_1 > 0$, which isn't explicitly stated in the question, otherwise you're taking AM-HM on a negative term. Jan 5 at 15:00