Difference between revisions of "1979 USAMO Problems/Problem 3"

Revision as of 14:12, 19 April 2014

Problem

$a_1, a_2, \ldots, a_n$ is an arbitrary sequence of positive integers. A member of the sequence is picked at random. Its value is $a$ . Another member is picked at random, independently of the first. Its value is $b$ . Then a third value, $c$ . Show that the probability that $a + b +c$ is divisible by $3$ is at least $\frac14$ .

First Hint

The given problem is equivalent to proving that $4(x^3 + y^3 + z^3 + 6xyz) \ge (x + y + z)^3$ .

Second Hint

Try proving this inequality for all positive real numbers x, y, z; not just positive integers.

Solution

This problem needs a solution. If you have a solution for it, please help us out by adding it.

@@ Line 7: / Line 7: @@
 The given problem is equivalent to proving that <math>4(x^3 + y^3 + z^3 + 6xyz) \ge (x + y + z)^3</math>.
 ==Second Hint==
-What do you do with ''homogenous'' inequalities? (A function <math>f</math> is ''homogenous'' of degree <math>d</math> if <math>f(kx) = k^df(x)</math> for all x in the domain of <math>f</math>.)
+Try proving this inequality for all positive real numbers x, y, z; not just positive integers.
 ==Solution==

1979 USAMO (Problems • Resources)
Preceded by Problem 2	Followed by Problem 4
1 • 2 • 3 • 4 • 5
All USAMO Problems and Solutions

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