từ giải thiết, ta có x(3x−2012)+y(3y−2012)+z(3z−2012)⩽2013<=>3(x2+y2+z2)⩽2013+2012(x+y+z)
Mà 3(x2+y2+x2)=(12+12+12)(x2+y2+z2)⩾(x+y+z)2
=>(x+y+z)2⩽2013+2012(x+y+z)
<=>(x+y+z)2−2012(x+y+z)−2013⩽0
<=>0⩽x+y+z⩽2013 (x,y,z dương )
A=x+y+z−(1x+1y+1z)
⩽x+y+z−9x+y+z(do1x+1y+1z⩾9x+y+z)
đặt t=x+y+z,A=t−9t=f(t)(0<t⩽2013)
Ta có f(t)=1+9t2>0∀t∈(0;2013]
f(t)max=f(2013)=2013−92013=40521602013 dấu "=" xảy ra <=>x=y=z=20133
Vậy maxA=40521602013<=>x=y=z=20133