Determine the largest natural number $m$ such that for each non negative real numbers $a_1 \ge a_2 \ge ... \ge a_{2014} \ge 0$ , it is true that $$\frac{a_1+a_2+...+a_m}{m}\ge \sqrt{\frac{a_1^2+a_2^2+...+a_{2014}^2}{2014}}$$
Problem
Source: INAMO Shortlist 2014 A5
Tags: algebra, inequalities
duck_master
22.05.2019 02:09
Solution to the problem as stated: Clearly for $m=2014$ the inequality holds. For $m=2015$ the inequality does not hold, since for any choice of $a_1, a_2, \dots, a_{2014}$, we can make $a_{2015}$ as small as we want to falsify it.
IndoMathXdZ
22.05.2019 07:26
parmenides51 wrote: Determine the largest natural number $m$ such that for each non negative real numbers $a_1 \le a2 \le ... \le a_{2014} \le 0$ , it is true that $$\frac{a_1+a_2+...+a_m}{m}\ge \sqrt{\frac{a_1^2+a_2^2+...+a_{2014}^2}{2014}}$$ Actually the corrected version should be : Determine the largest natural number $m$ such that for each non negative real numbers $a_1 \ge a_2 \ge \dots \ge a_{2014} \ge 0$, then the following holds. \[ \frac{a_1 + a_2 + \dots + a_m}{m} \ge \sqrt{\frac{a_1^2 + a_2^2 + \dots + a_{2014}^2}{2014}} \]
The answer is $m = 44$. To see that this is indeed the largest constant, notice that plug $(1,0,0,0\dots,0)$ to the inequality and we will get $m \le \sqrt{2014}$.
Now, to prove that $m = 44$ works, notice that
\[ 2014(a_1 + a_2 + \dots + a_{44})^2 \ge 1936(a_1^2 + a_2^2 + \dots + a_{2014}^2) \]Which is indeed true.
parmenides51
22.05.2019 07:32
I just replaced the incorrect $\le$ with the correct $\ge$, thanks for noticing