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 $b$?
 A: $a^2+ab+b^2= (a+ \frac{1}{2}b)^2 + \frac{3}{4}b^2$
One may prefer symmetry:
$2(a^2+ab+b^2)= a^2 +b^2+(a+b)^2$.
A: 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 $(a+\frac{3}{2}b)^2-(\frac{\sqrt5}{2}b)^2$.
A: Any second-degree polynomial with a negative discriminant always has the same sign. That follows from completing the square, as Khue did. In our case:
$$ 4(a^2+ab+b^2) = (2a+b)^2+\color{red}{3}b^2 $$
and that $\color{red}{3}$ is the opposite of the discriminant of $a^2+ab+b^2$, i.e. $1^2-4\cdot 1\cdot 1=-3.$
A: Or, one can argue via the sign of $ab$:
if
$ab \ge 0, \tag{1}$
then, since 
$a^2, b^2 \ge 0$ for all $a, b \in \Bbb R$,
$a^2 + b^2 + ab \ge 0; \tag{2}$
if
$ab < 0, \tag{3}$
then evidently
$ab = -\vert a \vert \vert b \vert, \tag{4}$
and we can without loss of generality assume that
$\vert a \vert \ge \vert b \vert, \tag{5}$
since the expression $a^2 + ab + b^2$ is invariant under the interchange $a \leftrightarrow b$.  Thus,
$a^2 + ab = \vert a \vert^2 - \vert a \vert \vert b \vert$
$= \vert a \vert (\vert a \vert - \vert b \vert) \ge 0, \tag{6}$
whence
$a^2 + ab + b^2 \ge 0, \tag{7}$
since $b^2 \ge 0$ for all $b \in \Bbb R$.
