H=4x2+2y2+7z2+2xy−4yz−6zx+2=(x2+y2+2xy)+3(x2−2zx+z2)+[(2z)2−4yz+y2]+2=(x+y)2+3(z−x)2+(y−2z)2+2≥2
Dấu = xảy ra tại x=y=z=0 J=2xx2+7x+4=2xx2+4x+4+2x3x=2x(x+2)2+23≥23 K=.....=x+1x2+4x+4+x+1x+1=x+1(x+2)2+1≥1 L=x2+x+12x2+2x+7
=> Lx2+Lx+L−2x2−2x−7=0 (L−2)x2+(L−2)x+(L−7)=0
Để x tồn tại <=> Δ=(L−2)2−4.(L−2)(L−7)≥0 (L−2)(L−2−4L+28)=(L−2)(−3L+26)≥0 (L−2)(3L−26)≤0
=> 2≤x≤326