The left side of the inequality can be factored into a product after grouping.

$$f(x)-f(y)=x^{2}-6 x+5-\left(y^{2}-6 y+5\right)=x^{2}-y^{2}-6(x-y)$$

from which

$$f(x)-f(y)=(x-y)(x+y-6)$$

Consider in the Cartesian coordinate system the lines given by the equations $x-y=0$ and $x+y-6=0$ (Figures 1a and 1b). These lines divide the plane into half-planes such that the sign of $x-y$ and $x+y-6$ is constant within each half-plane, while on the lines themselves, the value of $x-y$ and $x+y-6$ is naturally zero.

# 1988-02-073-1.eps

## Figure 1a

1988-02-073-2.eps

## Figure 1b

The desired set of points thus consists of the two lines and the common parts of the corresponding "positive" and "negative" half-planes (Figure 2), since a product is non-negative precisely when it is zero or when both factors have the same sign, i.e., both are positive or both are negative.

## 1988-02-074-1.eps

## Figure 2