a/(b+3c)+b/(8c+4b)+9c/(3a+2b)的最小值
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a/(b+3c)+b/(8c+4b)+9c/(3a+2b)的最小值
a/(b+3c)+b/(8c+4b)+9c/(3a+2b)的最小值
a/(b+3c)+b/(8c+4b)+9c/(3a+2b)的最小值
题目有问题应该是a/(b+3c)+b/(8c+4a)+9c/(3a+2b) 设b+3c=x,8c+4a=y,3a+2b=z,则 c=(8x-4z+3y)/48,b=(8x+4z-3y)/16,a=(4z-8x+3y)/24 所以原式变为(4z-8x+3y)/24x+(8x+4z-3y)/16y+9(8x-4z+3y)/48z即 z/6x+y/8x+x/2y+z/4y+3x/2z+9y/16z-61/48,利用平均值不等式 原式≥2[√(yz/48x^2)+√(xz/8y^2)+√(27xy/32z^2)]-61/48 不等式当且仅当x:y:z=3:8:6时成立 故原式≥2*(1/3+3/16+3/4)-61/48=61/48