Do x∈[−1;2]⇒x∈[−1;2]⇒(x+1)(x−2)≤0⇔x2≤x+2(x+1)(x−2)≤0⇔x2≤x+2
Tương tự: y2≤y+2y2≤y+2 ; z2≤z+2z2≤z+2
Cộng vế: x2+y2+z2≤x+y+z+6=6x2+y2+z2≤x+y+z+6=6 (đpcm)
Mặt khác x;y;z∈[−1;2]⇒(x+1)(y+1)(z+1)≥0x;y;z∈[−1;2]
⇔xyz+xy+yz+zx+x+y+z+1≥0
⇔xyz+xy+yz+zx+1≥0
⇔2xyz+2≥−2(xy+yz+zx)
⇔2xyz+2≥(x2+y2+z2)−(x+y+z)2
⇔2xyz+2≥x2+y2+z2