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.
If a=b=c,
which is a big "if",
then a = 2/3, (a+1/a)^2 = (13/6)^2 = 169/36
m/n = 169/12 = 14.08333 = Sum of the 3 terms.
m+n = 181
but; I have not proved all = 2/3 yields a minimum value for the expression.
Instead suppose a = 2/3, b = a+k, c=a-k, for some small k. Testing a few values suggests that a,b,c = 2/3 is at least a local minimum.
But what if some of a,b,c can be negative?
Some numeric brute force searching shows:
(a,b,c)= (1.40654, 1.40655, -0.81309)
to yield an amount for the expression:
13.141382 which is smaller than when a=b=c.
Try selecting fractions near the computed optima.
Let a = 7/5, b=7/5, c= -4/5
(these values do not produce a minimum, but they produce a number less than a=b=c=2/3)
(7/5+5/7)^2 + (7/5+5/7)^2 + (-4/5-5/4)^2
approx 13.142908
f(a,b,c) = 257601/19600 = m/n
m+n = 277201
Which probably is not the best solution.
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Posted by Larry
on 2025-02-22 08:50:59 |
thoughts
