Problem A. 478. (March 2009)

A. 478. If a₁,a₂,...,a_n are positive numbers with a₁+a₂+...+a_n=1, prove $a_1\cdot a_2^{2/3}+a_2\cdot a_3^{2/3}+\dots+a_{n-1}\cdot a_n^{2/3}<\frac37$ .

(5 pont)

Deadline expired on April 15, 2009.

Solution. Let

$S = a_1\cdot a_2^{2/3}+a_2\cdot a_3^{2/3}+\ldots+a_{n-1}\cdot a_n^{2/3}.$

Apply Hölder's innequality on the sequences (a_i^1/3) and (a_ia_i+1)^1/3 (i=1,2,...,n-1), with the expononts p=3 and q=3/2:

$S = \sum_{i=1}^{n-1} a_i^{1/3}\Big(a_i^{2/3}a_{i+1}^{2/3}\Big) \le \left(\sum_{i=1}^{n-1} \Big(a_i^{1/3}\Big)^3 \right)^{1/3} \left(\sum_{i=1}^{n-1}\Big((a_ia_{i+1})^{2/3}\Big)^{3/2}\right)^{2/3} =$

$= \left(\sum_{i=1}^{n-1} a_i \right)^{1/3} \left(\sum_{i=1}^{n-1} a_ia_{i+1}\right)^{2/3}.$

From the estimates $\sum_{i=1}^{n-1}a_i<\sum_{i=1}^na_i=1$ and

$\sum_{i=1}^{n-1} a_ia_{i+1} \le \left( \sum_{i\equiv0~(2)} a_i \right) \left( \sum_{j\equiv1~(2)} a_j \right) \le \left(\frac{ \left( \sum\limits_{i\equiv0~(2)} a_i \right) + \left( \sum\limits_{j\equiv1~(2)} a_j \right) }2\right)^2=\frac14$

we obtain

$S < \left(\frac14 \right)^{2/3} = \frac1{2\root3\of2} < 0,3969 < \frac37.$

Based on the solution of Nagy János

Statistics:

3 students sent a solution.

5 points: Nagy 235 János, Nagy 314 Dániel, Tomon István.

Problems in Mathematics of KöMaL, March 2009

3 students sent a solution.
5 points:	Nagy 235 János, Nagy 314 Dániel, Tomon István.

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