For real numbers x,y, and z, find the minimum value of 2x2+5y2+2z2+4xy−4yz−2z−2x.
Solution — click to reveal
We can write 2x2+5y2+2z2+4xy−4yz−2z−2x=(x2+4y2+z2+4xy−2xz−4yz)+(x2+z2+1+2xz−2x−2z+1)+y2−1=(x+2y−z)2+(x+z−1)2+y2−1.We see that the minimum value is −1, which occurs when x+2y−z=x+z−1=y=0, or x=21,y=0, and z=21.