Problem:
Let x1,x2,…,x6 be nonnegative real numbers such that x1+x2+x3+ x4+x5+x6=1, and x1x3x5+x2x4x6≥5401. Let p and q be positive relatively prime integers such that qp is the maximum possible value of x1x2x3+x2x3x4+x3x4x5+x4x5x6+x5x6x1+x6x1x2. Find p+q.
Solution:
Let r=x1x3x5+x2x4x6, and s=x1x2x3+x2x3x4+x3x4x5+x4x5x6+ x5x6x1+x6x1x2. By the Arithmetic Mean-Geometric Mean Inequality,
r+s=(x1+x4)(x2+x5)(x3+x6)≤(3(x1+x4)+(x2+x5)+(x3+x6))3=271,
with equality if and only if x1+x4=x2+x5=x3+x6=31. Therefore s≤271−5401=54019. In order for the inequalities to be equalities, it suffices to find values of the six variables for which r+s=271 and r=5401. The values of r=5401 and s=54019 can be achieved by taking x1=50x3= 103,x5=601,x2=31−x5=6019,x4=31x1=301, and x6=31−x3=301. Thus qp=54019, and p+q=559.
The problems on this page are the property of the MAA's American Mathematics Competitions