Re: [中學] 圓的滾動軌跡 & 幾題題目

[JohnMash (Paul)]

※ 引述《hotplushot (熱加熱)》之銘言：

: 6. a,b皆為大於0的數
:    a^3 + b^3 +4
:   _______________  此最小值為何
:    (a+1)(b+1)

   a^3+b^3 =  (a+b)(a^2-ab+b^2) = (a+b)[(a+b)^2-3ab]

   (a+1)(b+1) = ab + (a+b) + 1

   Fix a+b = K,

   then when ab has the maximum, (a+1)(b+1) has the maximum,

   and a^3+b^3 has the minimum,

   hence, A = (a^3+b^3+4)/[(a+1)(b+1)] has the minimum.

   However, ab has the maximum at a=b=K/2

   hence,

   min A = B = (2a^3+4)/(a+1)^2 = 2(a^3+2)/(a+1)^2

   a^3 + 2 = (a+1)^3 - 3a(a+1) + 1

   then

   B/2 = (a+1) - 3a/(a+1) + 1/(a+1)^2

       = (a+1) - 3 + 3/(a+1) + 1/(a+1)^2

   Let k = a+1

   and C/2 = k + 3/k + 1/k^2

   we will use the fact

   arithimetic mean is greater than or equal to geometric mean.

   Denote

   C/2 = (k/N)*N + 3/(Mk) *M + 1/(Lk^2) *L

   then C/2 >= (N+M+L) * [(k/N)^N * [3/(Mk)]^M * [1/(Lk^2)]^L]^{1/(N+M+L)}

   N=M+2L

   k/N = 3/(Mk) = 1/(Lk^2)

   3N = Mk^2, 3Lk^3=3N

   M=3Lk, N=Lk^3

   Lk^3 = 3Lk + 2L

     k^3 -3k - 2 = 0

     k=2, 3N=4M=24L

     N=8, M=6, L=1

   hence, min C/2 = 2 + 3/2 + 1/4 = 15/4

----------------------------------------

   The answer is 3/2


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◆ From: 123.194.224.241
※ 編輯: JohnMash        來自: 123.194.224.241      (04/30 23:51)
推 njru81l     :另解(a^3+1)(1+1)(1+1)>=(a+1)^3(廣義柯西)b^3+1亦同
→ njru81l     :則分子>=(a+1)^3 /4+(b+1)^3 /4+2
→ njru81l     :再算幾得上式 >= 3/2 (a+1)(b+1)
→ njru81l     :和分母相約，檢查等號成立可得最小值為 3/2
推 thisday     :推文這招讚^^