Mathematical and Physical Journal
for High Schools
Issued by the MATFUND Foundation
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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

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

and

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

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

### Statistics:

 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