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.
Various graphs give strong evidence the solution will involve two of the numbers equal.
f(x)=(x+1/x)2 + (x+1/x)2 + ((2-2x)+1/(2-2x))2
Not even WolframAlpha wants much to do with this.
It will give the derivative, the numerator of which is this:
24x^{7}-88x^{6}+120x^{5}-72x^{4}+7x^{3}+24x^{2}-24x+8
Asking for the roots of this gives a number like those broll and Larry suggested
x=1.406547155...
which is probably not rational, so
f(x)=13.141382207...
is probably also not rational.
Something's wrong.
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Posted by Jer
on 2025-02-22 13:06:45 |
my turn
