3未知數不等式與極限求極值

a + b + c = 1 => a = 1 - b - c ... (1)
a2 + b2 + c2 = 1 ... (2)
(1 - b - c)2 + b2 + c2 = 1
1 + b2 + c2 - 2b - 2c + 2bc + b2 + c2 - 1 = 0
2b2 - 2b + 2bc + 2c2 - 2c = 0
b2 + (c - 1)b + c(c - 1) = 0 ...(3) (b的二次元方程式)
判別式 (c - 1)2 - 4c(c - 1) >= 0 (因b為實數)
(c - 1)(c - 1 - 4c) >= 0
(c - 1)(3c + 1) <= 0
-1/3 <= c <= 1
同理 -1/3 <= a, b, c <= 1
a6, b6, c6 皆為正數故 0 <= a6, b6, c6 <= 1
(3) 的解分別為 [1 - c - √(1 - c)(3c + 1)] / 2及[1 - c + √(1 - c)(3c + 1)] / 2 ... (4)
從(1)可知當b = [1 - c - √(1 - c)(3c + 1)] / 2時a = [1 - c + √(1 - c)(3c + 1)] / 2, 相反亦然
代入S = a6 + b6 + c6 經反覆後得3c6 - 6c5 + 3c4 + 6c3 - 6c2 - 1
dS/dc = 6c(3c - 2)(c3 - c2 + 1) = 0
其實根為 c = 0, c = 2/3及c = -0.755(超出範圍)
代入(4)得(1, 0, 0) 及(2/3, -1/3, 2/3)對應S = 1及43/243
所以 43/243 <= a6 + b6 + c6 <= 1
當n趨於無限大時an + bn + cn的值取決於其絕對值為最大者
因a + b + c = 1 及 -1/3 <= a, b, c絕對值為最大者不可能是負數,而絕對值最大為a, b 或c = 1時,故此(an+1 + bn+1 + cn+1)/(an + bn + cn)的上限也是1
由(4)觀察所得,當c=2/3時a = 2/3, b= -1/3,若c略低於2/3時則a略大於2/3,可見max(a,b,c)的下限是2/3故此(an+1 + bn+1 + cn+1)/(an + bn + cn)的下限也是2/3

2009-09-05 17:48:00 補充：
亦可直接比較c, b = [1 - c - √(1 - c)(3c + 1)] / 2及a = [1 - c + √(1 - c)(3c + 1)] / 2,最大值者其值必大於2/3

由題目提供的條件
可得
-1/3<=a，b，c<=1

2009-09-07 20:25:27 補充：

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推論過程稍為繁瑣
有更好的作法請告知!




2009-09-07 20:26:56 補充：
末段「1>=a>=2/3」
應改為
1>=lim(n->∞) [a^(n+1)+b^(n+1)+c^(n+1)]/(a^n+b^n+c^n)>=2/3
很抱歉!

由題目提供的條件
可得
-1/3<=a，b，c<=1
002

亦可直接比較c, b = [1 - c - √(1 - c)(3c + 1)] / 2及a = [1 - c + √(1 - c)(3c + 1)] / 2,最大值者其值必大於2/3
a + b + c = 1 => a = 1 - b - c ... (1)

a2 + b2 + c2 = 1 ... (2)

(1 - b - c)2 + b2 + c2 = 1

1 + b2 + c2 - 2b - 2c + 2bc + b2 + c2 - 1 = 0

2b2 - 2b + 2bc + 2c2 - 2c = 0

b2 + (c - 1)b + c(c - 1) = 0 ...(3) (b的二次元方程式)

判別式 (c - 1)2 - 4c(c - 1) >= 0 (因b為實數)

(c - 1)(c - 1 - 4c) >= 0

(c - 1)(3c + 1) <= 0

-1/3 <= c <= 1

同理 -1/3 <= a, b, c <= 1

a6, b6, c6 皆為正數故 0 <= a6, b6, c6 <= 1

(3) 的解分別為 [1 - c - √(1 - c)(3c + 1)] / 2及[1 - c + √(1 - c)(3c + 1)] / 2 ... (4)

從(1)可知當b = [1 - c - √(1 - c)(3c + 1)] / 2時a = [1 - c + √(1 - c)(3c + 1)] / 2, 相反亦然

代入S = a6 + b6 + c6 經反覆後得3c6 - 6c5 + 3c4 + 6c3 - 6c2 - 1

dS/dc = 6c(3c - 2)(c3 - c2 + 1) = 0

其實根為 c = 0, c = 2/3及c = -0.755(超出範圍)

代入(4)得(1, 0, 0) 及(2/3, -1/3, 2/3)對應S = 1及43/243

所以 43/243 <= a6 + b6 + c6 <= 1

當n趨於無限大時an + bn + cn的值取決於其絕對值為最大者

因a + b + c = 1 及 -1/3 <= a, b, c絕對值為最大者不可能是負數,而絕對值最大為a, b 或c = 1時,故此(an+1 + bn+1 + cn+1)/(an + bn + cn)的上限也是1

由(4)觀察所得,當c=2/3時a = 2/3, b= -1/3,若c略低於2/3時則a略大於2/3,可見max(a,b,c)的下限是2/3故此(an+1 + bn+1 + cn+1)/(an + bn + cn)的下限也是2/3