Given:
a+b b+c c+a
--- + --- + --- = 9
c a b
and a
2 + b
2 + c
2 = 12.
Determine the minimum value of a+b+c.
At first I guessed that {1,2,3} was a solution to both conditions; but it is not.
All three of a,b,c cannot be equal or the first equation would be 6 instead of 9.
But speculate that a=b, c being some other value.
Sum to be minimized: 2a + c
This does not guarantee the solution will be minimal if a,b,c can be all distinct.
2a/c + 2(a+c)/a = 9
2a^2 + c^2 = 12
2a^2 + 2ac + 2c^2 = 9ac
2a^2 + 2c^2 = 7ac \ system of 2 eqn, 2 unknowns
2a^2 + c^2 = 12 /
Graph in Desmos using x for a, y for c
https://www.desmos.com/calculator/stskxzocgm
This shows an ellipse and 2 lines with 4 intersection points.
The point of intersection which looks like it will produce the most negative value for 2x+y is: (-2.395, -0.751)
Graphically from Desmos
so {a,b,c} ≈ {-2.395, -2.395, -0.751} sum ≈ -5.541
Which is close to what Wolfram Alpha shows
≈(-0.750534, -2.39131, -2.39131) sum ≈ -5.53315
Edited on February 22, 2024, 1:13 pm
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Posted by Larry
on 2024-02-22 11:27:44 |
speculation