(x ^ 2 + y ^ 2 + Z ^ 2) (a ^ 2 + B ^ 2 + C ^ 2) = (AX + by + CZ) ^ 2 to prove X / a = Y / b = Z / C

(x ^ 2 + y ^ 2 + Z ^ 2) (a ^ 2 + B ^ 2 + C ^ 2) = (AX + by + CZ) ^ 2 to prove X / a = Y / b = Z / C

Let this be a special case of Cauchy inequality
The proof is as follows:
Let f (T) = (x ^ 2 + y ^ 2 + Z ^ 2) T ^ 2 + 2 (AX + by + CZ) t + (a ^ 2 + B ^ 2 + C ^ 2)
deta=4(ax+by+cz)^2-4(x^2+y^2+z^2)(a^2+b^2+c^2)
Because (x ^ 2 + y ^ 2 + Z ^ 2) (a ^ 2 + B ^ 2 + C ^ 2) = (AX + by + CZ) ^ 2
So delta = 0
The equation: F (T) = 0 has and has only one real root
Sorting: F (T) = (x ^ 2 + y ^ 2 + Z ^ 2) T ^ 2 + 2 (AX + by + CZ) t + (a ^ 2 + B ^ 2 + C ^ 2) = 0
f(t)=(xt+a)^2+(yt+b)^2+(zt+c)^2=0
Since three nonnegative numbers add up to zero, all three must be zero,
When all three are equal to 0, then f (T) = 0
When: XT + a = 0 YT + B = 0 ZT + C = 0
f(t)=0 t=-a/x=-b/y=-c/z
So: X / a = Y / b = Z / C