Difference between revisions of "1987 IMO Problems/Problem 3"
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=== Solution 2 === | === Solution 2 === | ||
− | This | + | This solution is very similar to Solution 1 but uses a slightly different approach for the first part. It suffices to find <math>{a_i}</math> where <math>a_i</math> is positive. Let <math>f(a_1, a_2, ..., a_n) = a_1| x_1|+a_2 |x_2|+\cdots+a_n |x_n|\geq0</math>. |
By the [[Cauchy-Schwarz Inequality]], | By the [[Cauchy-Schwarz Inequality]], | ||
<cmath>\begin{split} | <cmath>\begin{split} |
Latest revision as of 04:36, 28 May 2023
Contents
Problem
Let be real numbers satisfying
. Prove that for every integer
there are integers
, not all 0, such that
for all
and
.
Solution
Solution 1
We first note that by the Power Mean Inequality, . Therefore all sums of the form
, where the
is a non-negative integer less than
, fall in the interval
. We may partition this interval into
subintervals of length
. But since there are
such sums, by the pigeonhole principle, two must fall into the same subinterval. It is easy to see that their difference will form a sum with the desired properties.
Solution 2
This solution is very similar to Solution 1 but uses a slightly different approach for the first part. It suffices to find where
is positive. Let
.
By the Cauchy-Schwarz Inequality,
This implies that
, and hence the codomain of
is
. The rest of the proof is similar to Solution 1.
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.
1987 IMO (Problems) • Resources | ||
Preceded by Problem 2 |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 4 |
All IMO Problems and Solutions |