It is known that x, y and Z are arbitrary real numbers, satisfying XY + YZ + ZX = 1. It is proved that XYZ (x + y + Z) ≤ 1 / 3

It is known that x, y and Z are arbitrary real numbers, satisfying XY + YZ + ZX = 1. It is proved that XYZ (x + y + Z) ≤ 1 / 3

(xy+yz+zx)^2=x^2y^2+y^2z^2+z^2x^2+2xyz(x+y+z)=1
X ^ 2Y ^ 2 + y ^ 2Z ^ 2 + Z ^ 2x ^ 2 = 1 / 3 (x ^ 2Y ^ 2 + y ^ 2Z ^ 2 + Z ^ 2x ^ 2) (1 + 1 + 1) ≥ 1 / 3 (XY + YZ + ZX) ^ 2 = 1 / 3 (Cauchy inequality)
So XYZ (x + y + Z) ≤ 1 / 3

It is proved that there is a constant C such that there is 1 + │ x + y + Z │ + │ XY + YZ + ZX │ + │ XYZ │ > C (│ x │ + │ y │ + │ Z │) for all real numbers x, y and Z It's a normal number. Type one less word

It can be proved that both sides take the square and C ^ 2 takes 1 / 3

If the real numbers x, y, Z satisfy the equations: xy x+2y＝1…(1) yz y+2z＝2…(2) zx z+2x＝3…(3) ， There is () A. x+2y+3z=0 B. 7x+5y+2z=0 C. 9x+6y+3z=0 D. 10x+7y+z=0

Y = x from (1) and (3)
x−2，z＝6x
x−3，
Therefore, X ≠ 0 is substituted into (2) to obtain x = 27
10，
So y = 27
7，z=-54．
It is verified that the solutions of this system satisfy the original equations
∴10x+7y+z=0．
Therefore, D

Given XYZ = 1, x + y + Z = 2, X2 + Y2 + Z2 = 16, find algebraic formula 1 xy+2z+1 yz+2x+1 The value of ZX + 2Y

xy+2z=xy+2（2-x-y）=（x-2）（y-2）
Similarly, YZ + 2x = (Y-2) (Z-2), ZX + 2Y = (Z-2) (X-2)
Original formula = Z − 2 + X − 2 + y − 2
(x−2)(y−2)(z−2)=(x+y+z)−6
xyz−2(xy+yz+xz)+4(x+y+z)−8=-4
thirteen

Given XYZ = 1, x + y + Z = 2, X2 + Y2 + Z2 = 16, find algebraic formula 1 xy+2z+1 yz+2x+1 The value of ZX + 2Y

xy+2z=xy+2（2-x-y）=（x-2）（y-2）
Similarly, YZ + 2x = (Y-2) (Z-2), ZX + 2Y = (Z-2) (X-2)
Original formula = Z − 2 + X − 2 + y − 2
(x−2)(y−2)(z−2)=(x+y+z)−6
xyz−2(xy+yz+xz)+4(x+y+z)−8=-4
thirteen

Given 2x + 2Y + xy = - 2,2y + 2Z + YZ = - 1,2z + 2x + ZX = 50, find the value of XYZ + 2 (XY + YZ + ZX) + 4 (x + y + Z) + 8. Clarify your ideas. Thank you

I didn't expect a simple way: the first expression gives x = - (2 + 2Y) / (2 + y), the second solution Z = - (2Y + 1) / (2 + y), and the third reduction is 54y ^ 2 + 216y + 210 = 0, that is, 6 (3Y + 5) (3Y + 7) = 0, so y = - 5 / 3 or y = - 7 / 3. These processes are very easy. With y, we can get x = 4, z = 7 or