Let a, B, x, y belong to R, and a ^ 2 + B ^ 2 = 1, x ^ 2 + y ^ 2 = 1, prove that the absolute value of AX + by is less than or equal to 1 In addition to using the mean inequality, there are other solutions

Let a, B, x, y belong to R, and a ^ 2 + B ^ 2 = 1, x ^ 2 + y ^ 2 = 1, prove that the absolute value of AX + by is less than or equal to 1 In addition to using the mean inequality, there are other solutions

The simplest is Cauchy inequality, which is the direct inference of Cauchy inequality
(ax+by)^2

Let a, B, x, y belong to R, and a ^ 2 + B ^ 2 = 1, x ^ 2 + y ^ 2 = 1, prove that the absolute value of AX + by is less than or equal to 1
Use the basic inequality.

Let a = cos α B = sin α cos ^ 2 α + sin ^ 2 α = 1
x=cosβ y=sinβ cos^2β+sin^2β=1
ax+by=cosαcosβ+sinαsinβ=cos(α-β)
|cos(α-β)|

When x equals 1 and Y equals 2, the absolute value of ax plus by is equal to 4. When x equals 1 inch, the absolute value of ax plus by is equal to 2
When x is equal to 1 and Y is equal to 2, the absolute value of ax plus by is equal to 4. When x is equal to 1 inch, the absolute value of ax plus by is equal to 2. Find the value of 2A plus B

Take x = 1. Y = 2 into ax + by = 4
A + 2B = 4
Then x = - 1. Y = 1 into ax + by = 2
- A + B = 2
So a + 2B = 4 or - 4
-A + B = 2 or - 2
Thus, four quadratic equations with two variables can be obtained: 1) a + 2B = 4 and - A + B = 2
2) A + 2B = 4 and - A + B = - 2
3) A + 2B = - 4 and - A + B = 2
4) A + 2B = - 4 and - A + B = - 2
The solvable solutions of a and B are: a = 0, B = 2
a=8/3,b=2/3
a=-8/3,b=-2/3
a=0,b=-2
The value of 2A + B is 2 or 6 or - 6 or - 2
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