Problem A. 439. (November 2007)

A. 439. Prove that

$\frac{a_1^2}{1-a_1} + \frac{a_2^2}{a_1-a_2} + \frac{a_3^2}{a_2-a_3} + \ldots + \frac{a_n^2}{a_{n-1}-a_n} > \frac12(a_1+2a_2+3a_3+\ldots+na_n)-1$

whenever $1>a_1>a_2>\ldots>a_n>0$ .

(5 pont)

Deadline expired on December 17, 2007.

Solution. Define also a₀=1, and let $L=\sum_{i=1}^n\frac{a_i^2}{a_{i-1}-a_i}$ , $R=\sum_{i=1}^nia_i$ , and $S=\sum_{i=1}^na_i$ . We have to prove $L>\frac12R-1$ .

Applying the Cauchy-Schwarz inequailty on the sequences $\frac{a_i}{\sqrt{a_{i-1}-a_i}}$ and $(2i+1)\sqrt{(a_{i-1}-a_i)}$ ,

$L = \sum_{i=1}^n\left(\frac{a_i}{\sqrt{a_{i-1}-a_i}}\right)^2 \ge \frac{\displaystyle\left(\sum_{i=1}^n \frac{a_i}{\sqrt{a_{i-1}-a_i}}\cdot(2i+1)\sqrt{(a_{i-1}-a_i)}\right)^2}% {\displaystyle\sum_{i=1}^n\left((2i+1)\sqrt{(a_{i-1}-a_i)}\right)^2} = \frac{\displaystyle\left(\sum_{i=1}^n(2i+1)a_i\right)^2}% {\displaystyle\sum\limits_{i=1}^n(2i+1)^2(a_{i-1}-a_i)} =$

$= \frac{(2R+S)^2}{\displaystyle 9a_0+\sum_{i=1}^{n-1}\big((2i+3)^2-(2i+1)^2\big)a_i-(2n+1)^2a_n} > \frac{(2R+S)^2}{\displaystyle 9+\sum_{i=1}^n(8i+8)a_i} = \frac{4R^2+4RS+S^2}{8R+8S+9} >$

$> \frac{4R^2+4RS}{8R+8S+9} = \frac12R-\frac{9R}{16R+16S+18} > \frac12R-\frac9{16} > \frac12R-1.$

Statistics:

6 students sent a solution.

5 points: Lovász László Miklós, Tomon István, Wolosz János.

3 points: 1 student.

0 point: 1 student.

Unfair, not evaluated: 1 solutions.

Problems in Mathematics of KöMaL, November 2007

6 students sent a solution.
5 points:	Lovász László Miklós, Tomon István, Wolosz János.
3 points:	1 student.
0 point:	1 student.
Unfair, not evaluated:	1 solutions.

Already signed up?
New to KöMaL?