If x, y>0 such that:
(x+y)(1/x+1/y)=5
Then, find the minimum value of:
(x
3+y
3)(1/x
3+1/y
3)
Start with the initial equation,
solve for y in terms of x,
find y^3,
plug into the expression.
(x+y)(1/x+1/y)=5
(x+y)^2 = 5xy
y^2 - 3xy + x^2 = 0
y = (3/2)x ± (1/2)√(9x^2 - 4x^2)
y = ((3/2) ± (1/2)√5)x
y = (3 ± √5)x/2
y^3 = (3 ± √5)^3 * x^3 / 8
if +
y^3 = (27 + 5√5 + 9√5(3 + √5)) * x^3 / 8
y^3 = (72 + 32√5) * x^3 / 8
y^3 = (9 + 4√5) * x^3
if -
y^3 = (27 - 5√5 - 9√5(3 - √5)) * x^3 / 8
y^3 = (9 - 4√5) * x^3
(x^3+y^3)(1/x^3+1/y^3) = M
M = (x^3+y^3)^2 / (xy)^3
M = (1 + 9±4√5)^2*x^6 / (9±4√5)*x^6
M = {(10+4√5)^2/(9+4√5) , (10-4√5)^2/(9-4√5)}
M ≈ {20, 20}
So for x,y that satisfy the original equation, the expression can only take on one value: 20.
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Posted by Larry
on 2025-03-28 11:54:11 |
solution
