Verification: (x ^ 3 / x + y) + (y ^ 3 / y + Z) + (Z ^ 3 / Z + x) is greater than or equal to (XY + YZ + ZX) / 2
It is proved that x, y, z > 0. According to Cauchy inequality, we get (x ^ 2 + y ^ 2 + Z ^ 2) (y ^ 2 + Z ^ 2 + x ^ 2) > = (XY + YZ + ZX) ^ 2 --- > x ^ 2 + y ^ 2 + Z ^ 2 > = XY + YZ + ZX. Therefore, we use Cauchy inequality to obtain [x (x + y) + y (y + Z) + Z (Z + x)] [x ^ 3 / (x + y) + y ^ 3 / (y + Z) + Z ^ 3 / (Z + x)] > = (x ^ 2 + y ^ 2 + Z ^ 2) ^ 2 --- > x
If the square of X + the square of Y + the square of Z - xy-yz-zx = 0, verify that x = y = Z
x ²+ y ²+ z ²= xy+yz+zx
x ²+ y ²+ z ²- xy-yz-xz=0
Double two
2x ²+ 2y ²+ 2z ²- 2xy-2yz-2xz=0
(x ²- 2xy+y ²)+ (y ²- 2yz+z ²)+ (z ²- 2xz+x ²)= 0
(x-y) ²+ (y-z) ²+ (z-x) ²= 0
The square is greater than or equal to 0, and the sum is equal to 0. If one is greater than 0, at least one is less than 0, which is not true. Therefore, all three are equal to 0
So X-Y = 0, Y-Z = 0, z-x = 0
x=y,y=z,z=x
So x = y = Z
Given x ^ 2-yz = y ^ 2-zx = Z ^ 2-xy, verify x = y = Z or x + y + Z = 0
Proof: because x ^ 2-yz = y ^ 2-zx = Z ^ 2-xy, x ^ 2-yz = y ^ 2-zx gets x ^ 2-y ^ 2 + ZX YZ = (x + y) * (X-Y) + Z (X-Y) = 0
That is, X-Y = 0 or x + y + Z = 0. Similarly, y ^ 2-zx = Z ^ 2-xy gets Y-Z = 0 or x + y + Z = 0
If x / 3 = Y / 1 = Z / 2 and XY + YZ + ZX = 99, then 2x ^ 2 + 12Y ^ 2 + 9z ^ 2=
Let x = 3k, y = k, z = 2K substitute: 3K ^ 2 + 2K ^ 2 + 6K ^ 2 = 99, so k = 3 or - 3, so x ^ 2 = 81, y ^ 2 = 9, Z ^ 2 = 36, so the original formula = 594
If x / 3 = Y / 1 = Z / 2 and XY + YZ + ZX = 99, then 2x ^ 2 + 12Y ^ 2 + 9z ^ 2 =?
x=3y z=2y
y=+ -3
x=+-9
z=+-6
2x^2+12y^2+9z^2=594
If X: Y: z = 3:1:2 and ZX XY YZ = 99, find the value of 2x ^ 2 + 12Y ^ 2 + 9z ^ 2
x: Y: z = 3:1:2, so x / 3 = Y / 1 = Z / 2, so x = 3Y, z = 2Y substitute ZX XY YZ = 996y ^ 2-3y ^ 2-2y ^ 2 = 99y ^ 2 = 99, so x ^ 2 = (3Y) ^ 2 = 9y ^ 2Z ^ 2 = (2Y) ^ 2 = 4Y ^ 2, so 2x ^ 2 + 12Y ^ 2 + 9z ^ 2 = 2 * 9y ^ 2 + 12Y ^ 2 + 9 * 4Y ^ 2 = 66y ^ 2 = 66 * 99 = 6534