Problem:
Let x1​,x2​,…,x6​ be nonnegative real numbers such that x1​+x2​+x3​+ x4​+x5​+x6​=1, and x1​x3​x5​+x2​x4​x6​≥5401​. Let p and q be positive relatively prime integers such that qp​ is the maximum possible value of x1​x2​x3​+x2​x3​x4​+x3​x4​x5​+x4​x5​x6​+x5​x6​x1​+x6​x1​x2​. Find p+q.
Solution:
Let r=x1​x3​x5​+x2​x4​x6​, and s=x1​x2​x3​+x2​x3​x4​+x3​x4​x5​+x4​x5​x6​+ x5​x6​x1​+x6​x1​x2​. 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​=31​x1​=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