標題:
~ 證不等式 ~
發問:
實數 x , y , z 滿足 x + y + z = 1 , 證明 : (x + 1/x)2 + (y + 1/y)2 + (z + 1/z)2 ≥ 33 + 1/3 更新: x , y , z 是正數。
最佳解答:
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Let a1 = x + 1/x, a2 = y + 1/y and a3 = z + 1/x. b1 = b2 = b3 = 1 Then by Cauchy-Schwarz's ineq.: [(x + 1/x)2 + (y + 1/y)2 + (z + 1/z)2](12 + 12 + 12) >= (x + y + z + 1/x + 1/y + 1/z)2 3[(x + 1/x)2 + (y + 1/y)2 + (z + 1/z)2] >= (1 + 1/x + 1/y + 1/z)2 Now using Arithmetic mean >= Harmonic mean: (x + y + z)/3 >= 3/(1/x + 1/y + 1/z) 1/3 >= 3/(1/x + 1/y + 1/z) 1/x + 1/y + 1/z >= 9 Hence: 3[(x + 1/x)2 + (y + 1/y)2 + (z + 1/z)2] >= (1 + 1/x + 1/y + 1/z)2 = 102 = 100 Finally (x + 1/x)2 + (y + 1/y)2 + (z + 1/z)2 >= 100/3 = 33 + 1/3 For equality to hold: In the CS ineq.: x + 1/x = y + 1/y = z + 1/z In the AM >= HM: x = y = z Hence equality holds when x = y = z = 1/3
其他解答:
has to sepcify x,y,z are positive. since x=1;y=1;z=-1 (x+1/x)^2+(y+1/y)^2+(z+1/z)^2 = 12
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