Question

For a1,......,an € R, a1<a2<....<an, show that
{n/(a1-a0)}+{(n-1)/(a2-a1)}+.......+ {1/(an- an-1)} >= summation of k=1 to n (k^2/ ak).

Answer 1

$$For \ a_1,\cdots, a_n \in \mathbb{R}, a_1 < a_2 <\dots < a_n$$

$$\frac {n}{a_1 - a_0} + \frac {n-1}{a_2-a_1} +\cdots + \frac {1}{a_n - a_n-1} \geq \sum_{k=1}^{n} \frac {k^2}{a_k}$$

After dividing both sides by right hand side

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

where

$$\sum_{k=1}^{n} \frac {\frac {n}{a_1 - a_0}}{\frac {k^2}{a_k}}= \frac {\frac {n}{a_1 - a_0}}{\frac {1^2}{a_1}} + \frac {\frac {n}{a_1 - a_0}}{\frac {2^2}{a_2}} + \dots + \frac {\frac {n}{a_1 - a_0}}{\frac {n^2}{a_n}}$$

$$\sum_{k=1}^{n} \frac {\frac {n-1}{a_2 - a_1}}{\frac {k^2}{a_k}}= \frac {\frac {n-1}{a_2 - a_1}}{\frac {1^2}{a_1}} + \frac {\frac {n-1}{a_2 - a_1}}{\frac {2^2}{a_2}} + \dots + \frac {\frac {n-1}{a_2 - a_1}}{\frac {n^2}{a_n}}$$

$$\sum_{k=1}^{n} \frac {\frac {1}{a_n - a_{n-1}}}{\frac {k^2}{a_k}}= \frac {\frac {1}{a_n - a_{n-1}}}{\frac {1^2}{a_1}} + \frac {\frac {1}{a_n - a_{n-1}}}{\frac {2^2}{a_2}} + \dots + \frac {\frac {1}{a_n - a_{n-1}}}{\frac {n^2}{a_n}}$$

To show that left hand side equal or bigger then 1, it's suffices to show that at least one of

 the summands bigger or equal to 1.

 One can always choose such a summand.

 Let it be :

$$\sum_{k=1}^{n} \frac {\frac {n}{a_1 - a_0}}{\frac {k^2}{a_k}}= \frac {\frac {n}{a_1 - a_0}}{\frac {1^2}{a_1}} + \frac {\frac {n}{a_1 - a_0}}{\frac {2^2}{a_2}} + \dots + \frac {\frac {n}{a_1 - a_0}}{\frac {n^2}{a_n}}$$

$$\frac{\frac{a}{b}}{\frac{c}{d}}= \frac{ad}{bc}$$

$$\Downarrow$$

$$\sum_{k=1}^{n} \frac {na_k}{(a_1 - a_0)k^2}= \frac {na_1}{(a_1 - a_0)1^2} + \frac {na_2}{(a_1 - a_0)2^2} + \dots + \frac {na_n}{(a_1 - a_0)n^2} \geq 1$$

And again, it's enough to show that at least one of the summands bigger or equal to 1

 for example first

$$\frac {na_1}{(a_1 - a_0)1^2} \geq 1$$

where

$$n \geq 1^2 \ and \ a_1 \geq (a_1 - a_0) \implies \frac {na_1}{(a_1 - a_0)1^2} \geq 1$$