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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