Solving: $(x-a)(x-b)=x-c$, $(x-c)(x-b)=x-a$ and $(x-c)(x-a)=x-b$ Given three distinct real numbers $a$, $b$, and $c$, show that at least two of the three following equations
$$(x-a)(x-b)=x-c$$
$$(x-c)(x-b)=x-a$$
$$(x-c)(x-a)=x-b$$
have real solutions.
My attempt: I tried to multiply all the equations side by side to obtain
$$(x-a)^2(x-b)^2(x-c)^2=(x-a)(x-b)(x-c)$$
then what next?

Let the three quadratics be
$$\begin{align}
p_1(x)&=(x-a)(x-b)-(x-c)\\
p_2(x)&=(x-b)(x-c)-(x-a)\\
p_3(x)&=(x-c)(x-a)-(x-b)\\
\end{align}$$
By symmetry, we may assume $a\le b\le c$.  But that implies
$$p_2(b)=-(b-a)\le0\quad\text{and}\quad p_3(c)=-(c-b)\le0$$
Thus $p_2$ and $p_3$ have at least one real root (since their graphs are upward-pointing parabolas), and therefore have two real roots (since they are quadratics).
The key here is to note that any permutation of $a$, $b$, and $c$ simply permutes the three quadratics.
