Problem:
For 1≤i≤215 let ai​=2i1​ and a216​=22151​. Let x1​,x2​,…,x216​ be positive real numbers such that
i=1∑216​xi​=1 and 1≤i<j≤216∑​xi​xj​=215107​+i=1∑216​2(1−ai​)ai​xi2​​
The maximum possible value of x2​=nm​, where m and n are relatively prime positive integers. Find m+n.
Solution:
For 1≤i≤216, let bi​=1−ai​​. Because ∑i=1216​ai​=1, it follows that ∑i=1216​bi2​=215. Also, if {xi​} is a sequence of positive real numbers with ∑i=1216​xi​=1, then ∑1≤i<j≤216​2xi​xj​=(∑i=1216​xi​)2−∑i=1216​xi2​=1−∑i=1216​xi2​.
Observe that for each i,(bi​xi​​)2−2152xi​​+(215bi​​)2=(bi​xi​​−215bi​​)2≥0, and thus summing over 1≤i≤216 yields ∑i=1216​((bi​xi​​)2−2152xi​​+(215bi​​)2)≥0. Because ∑i=1216​bi2​=215 and ∑i=1216​xi​=1, it follows that ∑i=1216​(bi​xi​​)2−2152​+21521​⋅215≥0, which is equivalent to 2151​≤∑i=1216​(bi​xi​​)2=∑i=1216​1−ai​xi2​​. Hence ∑1≤i<j≤216​2xi​xj​=1−∑i=1216​xi2​≤215214​+∑i=1216​1−ai​xi2​​−∑i=1216​xi2​=215214​+∑i=1216​1−ai​ai​xi2​​, so ∑1≤i<j≤216​xi​xj​≤215107​+∑i=1216​2(1−ai​)ai​xi2​​.
Equality occurs in this inequality if and only if for each i,bi​xi​​−215bi​​=0 or xi​=2151−ai​​. Therefore such a sequence {xi​} is unique and x2​=8603​. The requested sum is 3+860=863​.