<s>[INST] Prove that $a^2+ab+b^2\ge 0$ How to prove that $a^2+ab+b^2\ge 0$?
Obviously the squares are positive, but how can I be sure that $ab$ doesn't become too negative with a certain combination of $a$ and $b$?
 [/INST] There is yet another of seeing this.
Since $ab$ is between $-2ab$ and $2ab$, the quantity $a^2+ab+b^2$ must be between $a^2-2ab+b^2$ and $a^2+2ab+b^2$. But these two quantities obey
$$a^2-2ab+b^2=(a-b)^2\ge0 \tag{1}$$
and 
$$a^2+2ab+b^2=(a+b)^2\ge0 \tag{2}$$
with equality only if $a=b$ for (1), $a=-b$ for (2),or $a=b=0$ for (1) and (2).
Since $a^2+ab+b^2$ is inclusively between two non-negative quantities, it must be non-negative.
This motivates why the discriminant is so important - in cases such as $a^2+3ab+b^2=(a+b)^2+ab$, the quantity is not between those in (1) and (2), and completing the square will lead to a difference of squares rather than an addition of squares, i.e $(a+\frac{3}{2}b)^2-(\frac{\sqrt5}{2}b)^2$.
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