Mathematical and Physical Journal
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Problem A. 665. (March 2016)

A. 665. Let \(\displaystyle a_1,a_2,\ldots,a_n\) be distinct positive integers. Show that

\(\displaystyle 3\sum_{i=1}^{n}a_i^5+\bigg(\sum_{i=1}^{n}a_i\bigg)^{2} \ge 4\bigg(\sum_{i=1}^{n}a_i^3\bigg)\bigg(\sum_{i=1}^{n}a_i\bigg). \)

Proposed by: Mehtaab Sawhney, Commack, USA

(5 pont)

Deadline expired on April 11, 2016.

Solution (outline). We prove the inequality by induction on \(\displaystyle n\).

For \(\displaystyle n=1\) the statement is \(\displaystyle 3a_1^5+a_1^2\ge4a_1^4\) which is equivalent to \(\displaystyle a_1^2(a_1-1)(3a_1^2-a_1-1)\ge0\).

Now suppose the result holds for \(\displaystyle n=k\) then consider \(\displaystyle n=k+1\). By the symmetry we may assume \(\displaystyle a_{k+1}>a_{k}\ldots>a_1\). Using the inductive hypothesis it suffices to prove that

\(\displaystyle 3a_{k+1}^5+a_{k+1}\Big(a_{k+1}+2\sum_{i=1}^{k}a_i\Big)\ge 4a_{k+1}^4+4a_{k+1}\Big(\sum_{i=1}^{k}a_i^3\Big)+ 4a_{k+1}^3\Big(\sum_{i=1}^{k}a_i\Big). \)

This is equivalent to

\(\displaystyle 3a_{k+1}^5+a_{k+1}^2-4a_{k+1}^4\ge4a_{k+1}\Big(\sum_{i=1}^{k}a_i^3\Big)+\Big(4a_{k+1}^3-2a_{k+1}\Big)\Big(\sum_{i=1}^{k}a_i\Big). \)(1)

The numbers \(\displaystyle a_1,\ldots,a_k\) are distinct integers from the interval \(\displaystyle [1,a_{k+1}-1]\), so

\(\displaystyle \sum_{i=1}^{k} a_i \le \sum_{i=1}^{a_{k+1}-1} i = \frac{a_{k+1}(a_{k+1}-1)}2 \)


\(\displaystyle \sum_{i=1}^{k} a_i^3 \le \sum_{i=1}^{a_{k+1}-1} i^3 = \left(\frac{a_{k+1}(a_{k+1}-1)}{2}\right)^2. \)

By applying these estimates on the RHS of (1) we obtain an identity.


6 students sent a solution.
5 points:Baran Zsuzsanna, Glasznova Maja, Imolay András, Williams Kada.
4 points:Bodnár Levente.
2 points:1 student.

Problems in Mathematics of KöMaL, March 2016