证明题(详解)若正数a、b、c满足a/(b+c)=b/(a+c)-c/(a+b),求证:b/(a+c)≥(√17 - 1)/4
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证明题(详解)若正数a、b、c满足a/(b+c)=b/(a+c)-c/(a+b),求证:b/(a+c)≥(√17 - 1)/4
证明题(详解)
若正数a、b、c满足
a/(b+c)=b/(a+c)-c/(a+b),
求证:b/(a+c)≥(√17 - 1)/4
证明题(详解)若正数a、b、c满足a/(b+c)=b/(a+c)-c/(a+b),求证:b/(a+c)≥(√17 - 1)/4
证明:
b/(a+c)=c/(a+b)+a/(b+c)
令a+b=x,b+c=y,c+a=z,则
a=(x+z-y)/2,b=(x+y-z)/2,c=(y+z-x)/2,
从而原条件可化为:
(x+y)/z=(y+z)/x + (z+x)/y -1 = y/x +x/y + z/x +z/y -1≥ 2+ z/x + z/y -1 = z/x + z/y +1 ≥ 4z/(x+y) +1,
令(x+y)/z=t,则
t≥4/t +1
解得
t ≥ (1+√17)/2 或 t≤(1-√17)/2
所以
b/(a+c)=(x+y-z)/(2z)= t/2 - 1/2 ≥ (√17-1)/4
即b/(a+c)≥(√17 - 1)/4