At least two of the nonnegative real numbers $ a_1,a_2,...,a_n$ aer nonzero. Decide whether $ a$ or $ b$ is larger if $ a=\sqrt[2002]{a_1^{2002}+a_2^{2002}+\ldots+a_n^{2002}}$ and $ b=\sqrt[2003]{a_1^{2003}+a_2^{2003}+\ldots+a_n^{2003} }$
Problem
Source: Moldova NMO 2002 grade 11 problem nr.4
Tags: inequalities, AMC, USA(J)MO, USAMO
modularmarc101
05.11.2008 01:14
i would guess b is larger. just a guess
mathking123
05.11.2008 04:16
would the power mean inequality help?
DreamTeam
05.11.2008 14:06
modularmarc101 wrote: i would guess b is larger. just a guess Actually, the larger one is $ a$. mathking123 wrote: would the power mean inequality help? I don't think so.
TZF
06.11.2008 00:46
We can assume $ \{ a \}$ is non-increasing. Write $ \sqrt[k]{\sum_{i=1}^n a_i ^k }$ as $ a_1 \sqrt[k]{\sum_{i=1}^n \left( \frac{a_i}{a_1} \right)^k}$.
The summation under the radical cannot increase if we increase $ k$ because each of the $ n$ terms will either remain the same (if they are 1 or 0) or decrease.
Also, the summation inside the squareroot is at least $ 1$ because of the term $ (a_1/a_1)^k$ -- therefore, taking a higher root of it will necessarily cause it to decrease.
limes123
06.11.2008 00:51
I think this can be solved with help of induction. For n=2 (both nonzero) $ (a_1^{2002}+a_2^{2002})^{2003}>(a_1^{2003}+a_2^{2003})^{2002}$ can be proved with Newton's binomial theorem. I think that we can also use this theorem in inductive step.
t0rajir0u
06.11.2008 03:04
The relevant inequality is here. limes123's suggestion seems to work more or less fine.
kaiokan
09.11.2008 09:39
deleteddeleted
TZF
09.11.2008 19:51
Probably not. I don't know how the MNMO grades, but here is the scoring rubric for the USAMO: * 0 - No work, or completely trivial work * 1-2 - Progress on the problem, but not completely solved * 3-4 - All steps are present, but may lack clarity. (These scores are very rare.) * 5-6 - Complete solution with minor errors * 7 - Perfect solution Your response does not make progress towards a thorough solution, so it would receive 0 points if it were on the USAMO. Having said that: 1. Making appropriate guesses using context clues is very helpful in finding out the answer, which can help you get started towards a legitimate solution. 2. Considering appropriate cases is helpful in a short-answer test where proof is not required. 3. In a proof test, if this is what you come up with, and you have free time, its worth writing down to see if you get any points.
kjytay
22.07.2015 10:50
I think the following brute force method works: For fixed non-negative real numbers $a_1, \dots, a_n$, define the function $f(x) := \left( a_1^x + \dots + a_n^x \right)^{\frac{1}{x}}$. Directly differentiate the function $f$ and prove that the derivative is always non-positive. Hence, $a = f(2002) \geq f(2003) = b$.