Expanding this completing the square problem.

Question

The following is the final stage of a completed proof, which arrived at a completing the square problem that I'm having trouble figuring out the deriving steps. Could anyone help to show the steps of how the final result was arrived?

Very appreciative for any help!

when x1 and x2 not equal to zero.

Answer 1

You can just expand it to see that it works. The idea is to make a square that takes care of the cross term $-4x_1x_2$. That means you want $(ax_1-bx_2)^2$ with $2ab=4$. You also want $a^2\lt 9, b^2 \lt 6$ so you are left with positive squares. There are many choices. You could have chosen $a=2,b=1$ for example, getting $$9x_1^2+6x_2^2-4x_1x_2=(2x_1^2-x_2)^2+5x_1^2+3x_2^2\ge 0$$ just as well. You could also use $a=b=\sqrt 2$.

Answer 2

$9x_1^2+6x_2^2-4x_1x_2=(x_1^2-4x_1x_2+4x_2^2)+8x_1^2+2x_2^2$

Answer 3

To show that the expressin is non-negative, yo have to "hide" the possibly negative summand $-4x_1x_2$ in (the expansion of) a square. One could as well try to use $2(x_1-x_2)^2=2x_1^2-4x_1x_2+2x_2^2$ for this hiding, arriving at $$9x_1^2+6x_2^2-4x_1x_2=2(x_1-x_2)^2 +7x_1^2 +4x_2^2\ge 0,$$ for we are lucky enough to have sufficient leeway in the large coefficients $9$ and $6$ of the other summands.