推 sinker:恍然大悟~謝謝小帥!!!! 06/21 14:19
※ 引述《sinker (伸卡)》之銘言:
: 題目如下
: 設x、y為實數且2x+y=9
: 則x^2+y^2+2y+1之最小值為何?
: 我是用柯西去解答案為21,可是學生說答案是20,
: 故想請問版上強者此題之詳解為何?
: 謝謝大家
[x^2 + (y+1)^2] (2^2 + 1^2) ≧ (2x+y+1)^2 = 100
x^2+y^2+2y+1 ≧ 20
since x/2 = (y+1)/1 => x = 2y+2 => 4y+4+y=9 => (x, y) = (4, 1)
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