Each of a, b, and c is a real number such that a+b+c=2.
Suppose the minimum value of:
(a+1/a)2 + (b+1/b)2 + (c+1/c)2 is m/n, where gcd(m,n) =1.
Find m+n.
*** Adapted from a problem which appeared at Round 1 of Singapore Mathematical Olympiad Open, 2018.
Observation: (n+1/n) is at a minimum when n=1, so the theoretical minimum of the squared expression is 12.
But that is forbidden by the requirement that (a+b+c)=2.
So at least 2 terms of the squared expression must be greater than 4.
Note: (n/(n-1)+(n-1)/n)^2 tends to 4, but again the issue is (a+b+c)=2. Assume, then that c, say, is negative.
Assuming further by symmetry that a=b, computing small values of :
2(n/(n-1)+(n-1)/n)^2+(2-2(n/(n-1))+(1/(2-2(n/(n-1)))))^2
agrees that n must lie between 3 and 4.
n=3 gives 241/18 or 13.388 for the squared expression, while n=4 gives 107/8 or 13.378; values increase on either side of those numbers.
Searching for minima, the smallest value so far found for the squared expression is around 13.14, with very large m,n.
Edited on February 22, 2025, 4:12 am
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Posted by broll
on 2025-02-22 03:28:39 |
some thoughts
