Cho ba số thực x, y, z dương thỏa mãn: x+2y+3z =2

Do x+2y+3z=2 nên x=2−2y−3z2y=2−x−3z3z=2−x−2y, Khi đó:

xy+3z=xy+2−x−2y=xy−x−2y−2=xy−1−2y−1=x−2y−13yz+x=3yz+2−2y−3z=3yz−3z−2y−2=y−13z−23xz+4y=3xz+22−x−3z=3xz−6z−2x−4=3zx−2−2x−2=x−23z−2

Suy ra:
S=xyx−2y−1+3yzy−13z−2+3xzx−23z−2=x21−y2y2−x+2y1−3z3z21−y+x2−3z3z2−x≤12x21−y+2y2−x+122y2−3z+3z21−y+12x2−3z+3z2−x=12x21−y+2y2−x+2y2−3z+3z21−y+x2−3z+3z2−x=12x+3z21−y+2y+3z2−x+2y+x2−3z=122−2y2(1−y)+2−x2−x+2−3z2−3z=121+1+1=32

Hay S≤32⇒MaxS=32

Dấu”=” xảy ra ⇔x21−y=2y2−x2y2−3z=3z21−yx2−3z=3z2−x⇒2x−x2=4y−4y2=6z−9z2và x+2y+3z=2