Y = x (X-2) 2 / 2 (1 + x) 2, X ∈ (0, half), find the minimum value

Y = x (X-2) 2 / 2 (1 + x) 2, X ∈ (0, half), find the minimum value

y=(1+x)^2/[x(x-2)^2],x∈(0,1/2),
y'=[2(1+x)*x(x-2)^2-(1+x)^2*(3x^2-8x+4)]/[x(x-2)^2]^2
=(1+x)(2x^3-8x^2+8x
-3x^3+8x^2-4x
-3x^2+8x-4)/[x(x-2)^2]^2
=(x+1)(-x^3-3x+12x-4)/[x(x-2)^2]^2
=-(x+1)(x^2+5x-2)/[x^2*(x-2)^3]
=-(x+1)[x-(-5-√33)/2][x-(-5+√33)/2]/[x^2*(x-2)^3],
When x = (- 5 + √ 33) / 2, the minimum value of y = (3-5X) / [x (6-9x)] (using x ^ 2 = 2-5x)
=(3-5x)/(51x-18)
=(31-5√33)/(-291+51√33)
=(31-5√33)(291+51√33)/1152
=(606+126√33)/1152
=(101+21√33)/192.

Given that X and y are positive R, and 4x + 3Y = 1, what is the minimum value of 1 / x + 1 / y?

4x+3y=1
So 1 / x + 1 / y = (4x + 3Y) / x + (4x + 3Y) / y = 4 + 3Y / x + 3 + 4x / y = 7 + 3Y / x + 4x / Y > = 7 + radical (12) = 7 + 2 radical (3) (mean inequality is used here)
So the minimum value is: 7 + 2 radical (3)

x. If y is positive, 1 / x + 3 / y = 1 / 2, then the minimum value of 4x + 3Y?

1/x+3/y=1/2
SO 2 / x + 6 / y = 1
4x+3y=（4x+3y）（2/x+6/y）=8+24x/y+6y/x+18=26+24x/y+6y/x
>=24 + 2 radical (24 * 6) = 24 + 24 = 48
So the minimum value is 48, where 24x / y = 6y / X 4x & # 178; = y & # 178;

If 4x-3y = 0, then x + YY = 0___ ．

From 4x-3y = 0, we can get: xy = 34, x + YY = XY + 1 = 34 + 1 = 74