Re: [中學] 不等式一題
※ 引述《freePrester (Prester)》之銘言:
: 2 2 2
: 已知 x,y,z > 0 且 x + y + z = 1
: xy yz zx
: 試求 ---- + ---- + ---- 的最小值
: z x y
: 我知道答案是 √3 ,但我不知道要怎麼推出結論。
: 還請各位賜教。
(i)
(x^2 + y^2 + z^2)/3 = 1/3 ≧ (xyz)^(2/3)
=> (1/3)^(1/2) ≧ (xyz)^(1/3)
"="成立時
x^2 = y^2 = z^2 = 1/3
(ii)
[(xy/z)+(yz/x)+(zx/y)]/3 ≧ (xyz)^(1/3)
"="成立時
xy/z = yz/x = zx/y
=> x^2 = y^2 = z^2
由(i)(ii)可知
Min[(xy/z)+(yz/x)+(zx/y)] = 3*(1/3)^(1/2)
=> Min[(xy/z)+(yz/x)+(zx/y)] = sqrt[3]
--
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◆ From: 140.113.66.158
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感謝
※ 編輯: sulanpa 來自: 140.113.66.158 (12/29 08:53)
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"="成立時
(i) (ii) 條件相同
(xy/z)+(yz/x)+(zx/y)/3 = (1/3)^(1/2) = (xyz)^(1/3)
※ 編輯: sulanpa 來自: 140.113.66.158 (12/29 12:19)
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