a,b,c是不全相等的正數(shù),且abc=1,求證:1/a+1/b+1/c>√a+√b+√c
證：將不等式左邊變形為:
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),
又因為a,b,c不全相等,所以以上3式不能都取等號,
所以1/a+1/b+1/c>√(1/ab)+√(1/bc)+√(1/ac),又因為abc=1,
所以1/a+1/b+1/c>√a+√b+√c