If x > 0, Y > 0 and 2 / x + 1 / y = 2, then the minimum value of 3x + 2Y is 0

If x > 0, Y > 0 and 2 / x + 1 / y = 2, then the minimum value of 3x + 2Y is 0

2(3x+2y)
=(3x+2y)(2/x+1/y)
=6+2+(3x/y+4y/x)
>=8+4 √3
The minimum value of 3x + 2Y is 4 + 2 √ 3

If x, y, Z are positive real numbers and x-2y + 3Z = 0, then the minimum value of y2xz is ()
A. 4B. 3C. 2D. 1

∵ x-2y + 3Z = 0, ∵ y = x + 3z2, ∵ y2xz = x2 + 9z2 + 6xz4xz ≥ 6xz + 6xz4xz = 3, if and only if x = 3Z, take "=". So select B

If 3x + 2Y = 1, then the minimum value of x ^ 2 + y ^ 2
It is required to use the mean inequality ~ A ^ 2 + B ^ 2 to be greater than or equal to 2Ab

4/3

Let x, Y > 0, and 3x + 2Y = 1, find the minimum value of 1 / x + 1 / y
Please make it clear.

1/x+1/y
=（1/x+1/y）（3x+2y）
=3+3x/y+2y/x+2
=5+3x/y+2y/x
≥5+2√（3x/y*2y/x）
=5+2√6
If and only if 3x / y = 2Y / x, the minimum value is 5 + 2 √ 6
I hope I can help you. Happy study