消去 x=−12−y−zx=-\tfrac{1}{2}-y-zx=−21−y−z 得 (x−y)(x−z)+18=(2y+z+12)(y+2z+12)+18=18(4y+4z+1)2+(y+12)(z+12)(x-y)(x-z)+\tfrac{1}{8}=(2y+z+\tfrac{1}{2})(y+2z+\tfrac{1}{2})+\tfrac{1}{8}=\tfrac{1}{8}(4y+4z+1)^2+(y+\tfrac{1}{2})(z+\tfrac{1}{2})(x−y)(x−z)+81=(2y+z+21)(y+2z+21)+81=81(4y+4z+1)2+(y+21)(z+21)。因 y+12≥0y+\tfrac{1}{2}\ge0y+21≥0 且 z+12≥0z+\tfrac{1}{2}\ge0z+21≥0，右端非负，故 (x−y)(x−z)≥−18(x-y)(x-z)\ge-\tfrac{1}{8}(x−y)(x−z)≥−81。取等当且仅当 4y+4z+1=04y+4z+1=04y+4z+1=0 且 (y+12)(z+12)=0(y+\tfrac{1}{2})(z+\tfrac{1}{2})=0(y+21)(z+21)=0，即 x=−14x=-\tfrac{1}{4}x=−41 且 {y,z}={−12,14}\{y,z\}=\{-\tfrac{1}{2},\tfrac{1}{4}\}{y,z}={−21,41}。