What conditions must the numbers $a, b, c$ satisfy so that the expression $x^{2}+x y+y^{2}+a x+b y+c$ is positive for all pairs of numbers $(x, y)$?
The questionable expression can be rewritten in the form

$$
x^{2}+(y+a) x+\left(y^{2}+b y+c\right)
$$

This expression will be positive for all $(x, y)$ pairs if and only if (1) it is positive for all $x$ when $y$ is fixed at any arbitrary value. Since the coefficient of the $x^{2}$ term in (1) is positive, this will happen if and only if the discriminant of (1), which is $(y+a)^{2}-4\left(y^{2}+b y+c\right)$, is negative, i.e., if

$$
3 y^{2}+y(4 b-2 a)+\left(4 c-a^{2}\right)>0
$$

Thus, (1) will be positive for all $(x, y)$ pairs if and only if (2) holds for all $y$. This, in turn, will hold if the discriminant of (2) is negative, i.e.,

$$
a^{2}-a b+b^{2}<3 c
$$

which gives us the desired condition.

Nagy Attila (Györ, Czuczor G. Bencés Gymn., II. grade)

Second solution. By appropriately transforming four times the expression, we arrive at the following form:

$$
(2 x+y+a)^{2}+3\left(y+\frac{2 b-a}{3}\right)^{2}+\left(4 c-a^{2}-\frac{(2 b-a)^{2}}{3}\right)
$$

It is easily seen that this will be positive for all $(x, y)$ if and only if its third term is positive, from which we again obtain the (3) condition.

Remark. Our result means that the minimum of the function $x^{2}+x y+y^{2}+a x+b y$ is $\left(a b-a^{2}-b^{2}\right) / 3$.