If a > b > C and x > y > Z, how to prove ax + by + CZ > ay + BZ + CX?

If a > b > C and x > y > Z, how to prove ax + by + CZ > ay + BZ + CX?

ax+by+cz-ay-bz-cx
=a(x-y)+b(y-z)-c(x-z)
=a(x-y)+b(y-z)-c(x-y)-c(y-z)
=(a-c)(x-y)+(b-c)(y-z)
>0

Given a ^ 2 + B ^ 2 + C ^ 2 = 1, x ^ 2 + y ^ 2 + C ^ 2 = 9, find the maximum value of AX + by + CZ
The answer is 3

(a^2+b^2+c^2)+2m(ax+by+cz)+m^2(x^2+y^2+c^2)
=(a+mx)^2+(b+my)^2+(c+mz)^2>=0.
Let t = ax + by + CZ
That is: for any m, 1 + 2Mt + 9m ^ 2 > = 0 holds,
Because the opening is up, so the discriminant

Application of inequality: given a * a + b * B + C * C = 1, X * x + y * y + Z * z = 9. Then, what is the maximum value of AX + by + CZ?

∵a²+b²+c²=1,x²+y²+z²=9∴x²/9+y²/9+z²/9=1∴a²+b²+c²+x²/9+y²/9+z²/9=2∴(a²+x²/9)+(b²+y²/9)+(c²+z²...

If a ^ 2 + B ^ 2 + C ^ 2 = 1, x ^ 2 + y ^ 2 + Z ^ 2 = 9, then the maximum value of AX + by + CZ is______
It's better to explain in detail

From Cauchy inequality, there are: (a ^ 2 + B ^ 2 + C ^ 2) * (x ^ 2 + y ^ 2 + Z ^ 2) > = (AX + by + CZ) ^ 2, namely: (AX + by + CZ) ^ 2