(x + y + Z) (XY + YZ + XZ) - XY factorization

(x + y + Z) (XY + YZ + XZ) - XY factorization

（x+y+z）（xy+yz+xz）-xy=（x+y+z）xy+（x+y+z）（yz+xz）-xy
=(x + y + Z-1) XY + (x + y + Z) (x + y) Z i don't have time. If I have something to do, you can continue
One by one
Given X & # 178; + Y & # 178; + Z & # 178; - xy-yz-zx = 0, prove x = y = Z
X²+y²+z²-xy-yz-zx=0
2(X²+y²+z²-xy-yz-zx)=0
(x-y)²+(y-z)²+(z-x)²=0
So x = y = Z
X²+y²+z²-xy-yz-zx=0
2X²+2y²+2z²-2xy-2yz-2zx=0
(X-y)²+(y-z)²+(z-x)²=0
And: (X-Y) &# 178; ≥ 0, (Y-Z) &# 178; ≥ 0, (z-x) &# 178; ≥ 0
∴(X-y)²=(y-z)²=(z-x)²=0
∴X-y=y-z=z-x=0
∴x=y=z
X / (1 + x) + Y / (1 + y) + Z / (1 + Z) = 1 prove that x + y + Z is greater than or equal to 2 (XY + YZ + ZX)
　　∵x/(1+x)+y/(1+y)+z/(1+z)=1
　　∴（x（x+1）+y（y+1）+z（z+1））（x/(1+x)+y/(1+y)+z/(1+z)）≥（x+y+z）²
That is, X & # 178; + X + Y & # 178; + y + Z & # 178; + Z ≥ (x + y + Z) &# 178;
Expand the right side, that is, x + y + Z ≥ 2 (XY + YZ + ZX)
x. Y, Z ∈ (0,1), and X + y + Z = 2, prove 1
It is easy to know that any real number x, y, Z has X & sup2; + Y & sup2; + Z & sup2; ≥ XY + YZ + ZX. (x + y + Z) & sup2; = x & sup2; + Y & sup2; + Z & sup2; + 2 (XY + YZ + ZX) ≥ 3 (XY + YZ + ZX) ｜ XY + YZ + ZX ≤ (x + y + Z) & sup2 / / 3 = 4 / 3. Further prove the left. XY + YZ + ZX = XY + (x + y) (2-x-y) = - X & sup2