Re: [中學] 請教一題不等式
※ 引述《iamOrz (I am Orz)》之銘言：
: a,b是正實數,並且滿足a+b=1.
: 1 1
: 證明 (a + ---------)*(b + ---------) ≧ 25/4
: 2*(b^2) 2*(a^2)
: 請指教,謝謝
柯西不等式: ( a1^2 + a2^2 )( b1^2 + b2^2 ) >= ( a1*b1+a2*b2 )^2
令: a1 = sqrt(a), a2 =1/(sqrt(2)*b);
b1 = sqrt(b), b2= 1/(sqrt(2)*a). 代入上式
可得:
( a + 1/(2*b^2) )( b + 1/(2*a^2) ) >= ( sqrt(a*b) + 1/(2*a*b) )^2
上式當 a1/b1 = a2/b2 時(兩向量平行)等號成立:
即 sqrt(a)/sqrt(b) = a/b 成立時有最小值
又 a + b = 1, 是故 a = b =1/2 時, 左式有最小值( 1/2 + 2 )^2 = 25/4
( a + 1/(2*b^2) )( b + 1/(2*a^2) ) >= 25/4 得證
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◆ From: 125.231.33.179
※ 編輯: Rasin 來自: 125.231.33.179 (09/20 22:55)
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