已知x、y、z均為正數,求證:33(1x+1y+1z)≤1x2+1y2+1z2.

已知x、y、z均為正數,求證:33(1x+1y+1z)≤1x2+1y2+1z2.

證明:由柯西不等式得(12+12+12)(1x2+1y2+1z2)≥(1x+1y+1z)2…(5分)則3×1x2+1y2+1z2≥1x+1y+1z,即33(1x+1y+1z)≤1x2+1y2+1z2…(10分)