X-Y = 6, Y-Z = 5, find the square of X + the square of Y + the square of z-xy-yz-zx

X-Y = 6, Y-Z = 5, find the square of X + the square of Y + the square of z-xy-yz-zx

From X-Y = 6 and Y-Z = 5, x-z = 11
(x-y)2=x2+y2-2xy=36
(y-z)2=y2+z2-2yz=25
(x-z)2=x2+z2-2xz=121
Add the above three equations
X2 + y2-2xy + Y2 + z2-2yz + x2 + z2-2xz = 182
Divide by two
X2 + Y2 + Z2 XY YZ ZX = 91

Given 1 / x + 1 / y = 1 / 2, 1 / y + 1 / z = 1 / 3, 1 / Z + 1 / x = 1 / 4, find the value of XYZ / XY + YZ + ZX

1/X+1/Y=1/2,
1/Y+1/Z=1/3,
1/Z+1/X=1/4,
1/X+1/Y+1/Z=1/2*(1/2+1/3+1/4)=13/24
XYZ/XY+YZ+ZX=1/(1/Z+1/X+1/Y)=24/13

If the real numbers x, y, Z satisfy x ^ 2 + y ^ 2 + Z ^ 2 = 1 and X + y + Z = 0, the value range of the real number XY + YZ + ZX is

Suppose x, y and Z are positive numbers, because x2 + Y2 + Z2 = 1, (x2 + Y2) + (x2 + Z2) + (Y2 + Z2) = 2, and because x2 + Y2 > = 2XY, X2 + Z2 > = 2XZ, Y2 + Z2 > = 2yz, (x2 + Y2) + (x2 + Z2) + (Y2 + Z2) > = 2 (XY + XZ + YZ) that is, 2 > = 2 (XY + XZ + YZ) so XY + XZ + YZ

If the real numbers x, y, Z satisfy x ²+ y ²+ z ²＝ 1, and X + y + Z ≠ 0, then the value range of XY + YZ + ZX is

(x + y) 2 + (y + Z) 2 + (x + Z) 2 = 2 (x2 + Y2 + Z2 + xy = YZ = ZX) = - z) 2 + (- x) 2 + (- y) 2 = x2 + Y2 + Z2 = 1, that is, 2 (1 + XY + YZ + ZX) = 1, XY + YZ + ZX = - 1 / 2. (because I don't know how to get the square, only one of them is 2, please understand)

XYZ is a positive real number. Find the maximum value of XY + YZ / x ^ 2 + y ^ 2 + Z ^ 2

Mean inequality, x, y, Z are positive real numbers with
X ^ 2 + (y ^ 2) / 2 ≥ XY √ 2. ① (equal sign holds) x ^ 2 = (y ^ 2) / 2
(y ^ 2) / 2 + Z ^ 2 ≥ YZ √ 2. ② (equal sign holds) (y ^ 2) / 2 = Z ^ 2
① + ②
x^2+y^2/2+y^2/2+z^2≥xy√2+yz√2=√2(xy+yz)
therefore
(xy+yz)/(x^2+y^2+z^2)≤1/√2=(√2)/2
Therefore, if and only if x ^ 2 = (y ^ 2) / 2 = Z ^ 2, that is, x = (√ 2) y / 2 = Z, (XY + YZ) / (x ^ 2 + y ^ 2 + Z ^ 2) obtains the maximum value (√ 2) / 2

It is proved that there is a normal number C, so that for all real numbers x, y, Z, there is 1 + | x + y + Z | + | XY + YZ + ZX | + | XYZ | > C (| x | + | y | + | Z |)

Because 1 + | x + y + Z | + | XY + YZ + ZX | + | XYZ | > 0
And C (|x| + |y| + |z|) > 0
Take C