Mathematical and Physical Journal
for High Schools
Issued by the MATFUND Foundation
Already signed up?
New to KöMaL?

Problem A. 478. (March 2009)

A. 478. If a1,a2,...,an are positive numbers with a1+a2+...+an=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 (ai1/3) and (aiai+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)
\left(\sum_{i=1}^{n-1} \Big(a_i^{1/3}\Big)^3 \right)^{1/3}

= \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)
\left( \sum\limits_{i\equiv0~(2)} a_i \right) +
\left( \sum\limits_{j\equiv1~(2)} a_j \right)

we obtain

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

Based on the solution of Nagy János


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

Problems in Mathematics of KöMaL, March 2009