Xét $P=x+y+z-xyz$Ta phải CM:$P\leq2$
Ta có:
$P=x(1-yz)+(y+z)\leq \sqrt{x^{2}+(y+z)^{2}}.\sqrt{(1-yz)^{2}+1^{2}} $
= $\sqrt{2+2yz} . \sqrt{2-2yz+(yz)^{2}}$(BĐT Bunhiacopxki)
Đặt $t=yz \leq \frac{y^{2}+z^{2}}{2} \leq \frac{x^{2}+y^{2}+z^{2}}{2} =1 $
$\Rightarrow$ $t^{3} \leq t^{2} $(*)
$P=\sqrt{(2+2t)(2-2t+t^{2})} = \sqrt{2(t^{3}-t^{2}+2)} \leq 2$(do(*))
$\Rightarrow$đpcm
Dấu''='' xra$\Leftrightarrow$ $x=0$ & $y=z=1$