已知a、b、c为不全相等的正数,且abc=1,求证1/a+1/b+1/c>根号a+根号b+根号c

Question

已知a、b、c为不全相等的正数,且abc=1,求证1/a+1/b+1/c>根号a+根号b+根号c

Answer 1

a,b,c是不全相等的正数,且abc=1,求证:1/a+1/b+1/c>√a+√b+√c 证：将不等式左边变形为:1/a+1/b+1/c=1/2(1/a+1/a)+1/2(1/b+1/b)+1/2(1/c+1/c)= 1/2(1/a+1/b)+1/2(1/b+1/c)+1/2(1/a+1/c),由均值不等式得:1/2(1/a+1/c)≥√(1/ab) 1/2(1/b+1/c)≥√(1/bc) 1/2(1/a+1/c)≥√(1/ac),又因为a,b,c不全相等,所以以上3式不能都取等号,所以1/a+1/b+1/c>√(1/ab)+√(1/bc)+√(1/ac),又因为abc=1,所以1/a+1/b+1/c>√a+√b+√c