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

(Hint)
(Second Hint)
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The given problem is equivalent to proving that <math>4(x^3 + y^3 + z^3 + 6xyz) \ge (x + y + z)^3</math>.
 
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==
 
==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>.)
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Try proving this inequality for all positive real numbers x, y, z; not just positive integers.
  
 
==Solution==
 
==Solution==

Revision as of 15:12, 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.

See Also

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

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