Hellenic Mathematical Olympiad

The discriminant equals to $$\begin{array}{rl}\Delta & =16(a+b+c{)}^2-48(ab+bc+ca)\\ & =8((a-b{)}^2+(b-c{)}^2+(c-a{)}^2).\end{array}$$ Since two of the numbers are at least $\frac{1}{2\sqrt{2}}$ apart, the square of their difference will be greater than $1/8$, so $\Delta >1$. Since the discriminant is positive, the trinomial has two real roots, let ${\rho }_1>{\rho }_2$ between which the sign of the trinomial is negative. In addition, we have: $${\rho }_1-{\rho }_2=\frac{4(a+b+c)+\sqrt{\Delta }}{2}-\frac{4(a+b+c)-\sqrt{\Delta }}{2}=\sqrt{\Delta }>1,$$ so between ${\rho }_1,{\rho }_2$ there is an integer, say $x$, which makes the given trinomial negative according to the above.

Solution 2: Consider the function $$\begin{array}{rl}f(x)& =x^2-4(a+b+c)x+12(ab+bc+ca)\\ & =x(x-4(a+b+c))+12(ab+bc+ca).\end{array}$$ Observe that $$\begin{array}{rl}f(2(a+b+c))& =-4(a+b+c{)}^2+12(ab+bc+ca)\\ & =-2((a-b{)}^2+(b-c{)}^2+(c-a{)}^2)<0.\end{array}$$ The graph of $f$ is a parabola, which is convex and has the vertical line $x=2(a+b+c)$ as its axis of symmetry. Therefore $$f(2(a+b+c)+\frac{1}{2})=f(2(a+b+c)-\frac{1}{2}).$$

$$\begin{array}{rl}& f(2(a+b+c)+\frac{1}{2})\\ & =2(a+b+c)+\frac{1}{2}(-2(a+b+c)+\frac{1}{2})+12(ab+bc+ca)\\ & =\frac{1}{4}-4(a+b+c{)}^2+12(ab+bc+ca)\\ & =\frac{1}{4}-2((a-b{)}^2+(b-c{)}^2+(c-a{)}^2)\\ & =\frac{1-8((a-b{)}^2+(b-c{)}^2+(c-a{)}^2)}{4}<0,\end{array}$$ since $(a-b{)}^2+(b-c{)}^2+(c-a{)}^2>{\left(\frac{1}{2\sqrt{2}}\right)}^2=\frac{1}{8}$. This means that the trinomial has negative sign on the interval $[2(a+b+c)-\frac{1}{2},2(a+b+c)+\frac{1}{2}]$, which has length 1. Therefore, there is an integer on that interval having the desired property.