已知a^3-3a^2+5a=1,b^3-3b^2+5b=5,求a+b的值?已知a1,a2,a3,......,an都为正数,且和为1,求证:a1^2/(a1+a2)+a2^2/(a2+a3)+......+an^2/(an+a1)>=1/2.
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![已知a^3-3a^2+5a=1,b^3-3b^2+5b=5,求a+b的值?已知a1,a2,a3,......,an都为正数,且和为1,求证:a1^2/(a1+a2)+a2^2/(a2+a3)+......+an^2/(an+a1)>=1/2.](/uploads/image/z/6870718-46-8.jpg?t=%E5%B7%B2%E7%9F%A5a%5E3-3a%5E2%2B5a%3D1%2Cb%5E3-3b%5E2%2B5b%3D5%2C%E6%B1%82a%2Bb%E7%9A%84%E5%80%BC%3F%E5%B7%B2%E7%9F%A5a1%2Ca2%2Ca3%2C......%2Can%E9%83%BD%E4%B8%BA%E6%AD%A3%E6%95%B0%EF%BC%8C%E4%B8%94%E5%92%8C%E4%B8%BA1%EF%BC%8C%E6%B1%82%E8%AF%81%EF%BC%9Aa1%5E2%2F%28a1%2Ba2%29%2Ba2%5E2%2F%28a2%2Ba3%29%2B......%2Ban%5E2%2F%28an%2Ba1%29%3E%3D1%2F2.)
已知a^3-3a^2+5a=1,b^3-3b^2+5b=5,求a+b的值?已知a1,a2,a3,......,an都为正数,且和为1,求证:a1^2/(a1+a2)+a2^2/(a2+a3)+......+an^2/(an+a1)>=1/2.
已知a^3-3a^2+5a=1,b^3-3b^2+5b=5,求a+b的值?
已知a1,a2,a3,......,an都为正数,且和为1,求证:a1^2/(a1+a2)+a2^2/(a2+a3)+......+an^2/(an+a1)>=1/2.
已知a^3-3a^2+5a=1,b^3-3b^2+5b=5,求a+b的值?已知a1,a2,a3,......,an都为正数,且和为1,求证:a1^2/(a1+a2)+a2^2/(a2+a3)+......+an^2/(an+a1)>=1/2.
第一题:
∵ x³-3x²+5x=(x-1)³+2(x-1)+3
∴ (a-1)³+2(a-1)+3=1 …… ①
(b-1)³+2(b-1)+3=5 …… ②
①+②得:令a-1 = m ; b-1 = n
m³+2m+n³+2n=0
(m+n)[m²-mn+n²]+2(m+n)=0
(m+n)[m²-mn+n²+2]=0
∵m²-mn+n² > 0 恒成立
∴ m+n=0
即:a-1+b-1= 0
∴ a+b = 2
第二题
由柯西不等式
[(a1+a2)+(a2+a3)+(a3+a4)+.+(an+a1)]*[a1^2/(a1+a2)+a2^2/(a2+a3)+……+an^2/(an+a1)]
≥ [√(a1^2)+√(a2^2)+...+√(an^2)]^2
即:
2*[a1^2/(a1+a2)+a2^2/(a2+a3)+……+an^2/(an+a1)]
≥(a1+a2+……+an)^2=1
所以:
a1^2/(a1+a2)+a2^2/(a2+a3)+……+an^2/(an+a1)≥1/2