Given (4x ^ 2 + 11xy-3y ^ 2) / (x + 3Y), find the minimum value of Y ^ 2 + 8x, good to add, fast Correct: given (4x ^ 2 + 11xy-3y ^ 2) / (x + 3Y) = 0, find the minimum value of Y ^ 2 + 8x

Given (4x ^ 2 + 11xy-3y ^ 2) / (x + 3Y), find the minimum value of Y ^ 2 + 8x, good to add, fast Correct: given (4x ^ 2 + 11xy-3y ^ 2) / (x + 3Y) = 0, find the minimum value of Y ^ 2 + 8x

(4x^2+11xy-3y^2)/（x+3y）=0
4X ^ 2 + 11xy-3y ^ 2 = 0 and X + 3Y = 0
(4x-y)(x+3y)=0
4x-y=0
y=4x
therefore
y^2+8x
=16x^2+8x
=(4x)²+8x+1-1
=(4x+1)²-1≥-1
So the minimum value of Y ^ 2 + 8x is - 1

The square of 3x (X-Y) + 6x (Y-X)

3x square (X-Y) + 6x (Y-X) = 3x square (X-Y) - 6x (X-Y)
=(the square of 3x-6x) (X-Y)
=3x(x-2)(x-y)

When y is equal to the opening of negative 3x square + 6x-4, y increases with the increase of X

Y is equal to the opening of minus 3x square + 6x-4（