Problem 1242

The first solution. Let $f_{1}(x)=(x-b)(x-c)-(x-a)$, $f_{2}(x)=(x-c)(x-a)-(x-b)$, and $f_{3}(x)=(x-a)(x-b)-(x-c)$. Suppose the statement of the problem is false, that is, the maximum of one of these functions is a root. Then, two of them, say $f_{1}$ and $f_{2}$, do not have roots. Since the leading coefficients of these quadratic polynomials are positive, we get that $f_{1}(x)>0$ and $f_{2}(x)>0$ for all $x$. However, the polynomial

$$\begin{gathered} f_{1}(x)+f_{2}(x)=(x-c)(x-b+x-a)-(x-a+x-b)= \\ =(2 x-a-b)(x-c-1) \end{gathered}$$

has, for example, a root $x_{0}=c+1$; hence, it is false that $f_{1}\left(x_{0}\right)>0$ and $f_{2}\left(x_{0}\right)>0$. Contradiction.

The second solution. Let, for definiteness, $a < b < c$. Consider the functions $f_{a}(x)=(x-b)(x-c)-(x-a)$, $f_{b}(x)=(x-c)(x-a)-(x-b)$, and $f_{c}(x)=(x-a)(x-b)-(x-c)$. The function $f_{a}(x)$ is a parabola opening upwards, and its vertex lies above the x-axis: $f_{a}(a)=0$, $f_{a}(b)>0$, and $f_{a}(c)>0$. The function $f_{b}(x)$ is also a parabola opening upwards, and its vertex lies below the x-axis: $f_{b}(a)<0$, $f_{b}(b)=0$, and $f_{b}(c)>0$. Therefore, the equation $f_{a}(x)=f_{b}(x)$ has a root on the interval $[a, b]$ (see Fig. 1).

Similarly, the function $f_{c}(x)$ is a parabola opening upwards, and its vertex lies below the x-axis: $f_{c}(a)>0$, $f_{c}(b)<0$, and $f_{c}(c)=0$. The function $f_{a c}(x)=(x-b)(x-c)-(x-a)$ is a parabola opening upwards, and its vertex lies above the x-axis: $f_{a c}(a)=0$, $f_{a c}(b)<0$, and $f_{a c}(c)>0$. Therefore, the equation $f_{a c}(x)=f_{b}(x)$ also has a root on the interval $[a, b]$ (see Fig. 2).

The third solution. As in the first solution, suppose that $f_{1}(x)$ and $f_{2}(x)$ do not have roots. Then their discriminants are negative, that is, $(b+c+1)^{2}<4(b c+a)$ and $(c+a+1)^{2}<4(c a+b)$. These inequalities can be rewritten as $(b-c-1)^{2}<4 a-4 b$ and $(c-a+1)^{2}<4 b-4 a$. Therefore, both numbers on the right-hand sides are positive; however, their sum is zero. Contradiction.