高中數學問題最小值

3x - 4y = 20
x = (4y + 20)/3

代入得：
(x - 1)² + (y + 3)² + 1
= {[(4y + 20)/3] - 1}² + (y + 3)² + 1
= [(4y/3) + (17/3)]² + (y + 3)² + 1
= (16/9)y² + (136/9)y + (289/9) + y² + 6y + 9 + 1
= (25/9)y² + (190/9)y + (379/9)
= (25/9)[y² + (38/5)y] + (379/9)
= (25/9)[y² + (38/5)y + (38/10)²] - (25/9)(38/10)² + (379/9)
= (25/9)[y + (38/10)]² - (25/9)(38/10)² + (379/9)
= (25/9)[y + (38/10)]² - (361/9) + (379/9)
= (25/9)[y + (38/10)]² + (18/9)
= (25/9)[y + (38/10)]² + 2

由於當 y 為任何實數時， (25/9)[y +(38/10)]² ≥ 0
故此 (x - 1)² + (y + 3)² + 1 = (25/9)[y + (38/10)]² + 2 ≥ 2

所以 (x - 1)² + (y + 3)² + 1 的最小值 = 2

用柯西不等式

[(x-1)^2+(y+3)^2]*[3^2+(-4)^2]>=[3(x-1)+(-4)(y+3)]^2

==>[(x-1)^2+(y+3)^2]*25>=(3x-4y-15)^2

==>[(x-1)^2+(y+3)^2]*25>=(20-15)^2

==>[(x-1)^2+(y+3)^2]>=1

[(x-1)^2+(y+3)^2]+1>=1+1=2

其最小值為2

由(x-1)^2+(y+3)^2+1可知，求(x,y)到(1,-3)距離的平方再加1
故最小值為點(1,-3)到直線3x-4y=20的距離，點到直線公式可求得距離平方為1，所求為1+1=2
min=2