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?
 A: 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.
A: Let (x-a) (x-b) / (x-c) = y
Solving, $ x^2 -(a+b+y)x + ab + cy = 0 $
D>0
$(a+b+y)^2 -4(ab+cy) > 0 $ 
$a^2 + b^2 + y^2 + 2ab + 2by + 2ay - 4ab - 4cy >0$
$ y^2 + (2b+2a-4c)y + (a-b)^2 > 0 $
Now for this Quadratic, D<0 gives the quadratic the whole Real values as range.
$ 4(b+a-2c)^2 - 4(a-b)^2 < 0$
$(a-c)(b-c)<0$
$Either, a<c<b, or ,b<c<a.$ Which means, when c is the middle value, it will have a solution for y=1.
Now when in the RHS, it is the smallest value(assume a here). The value of LHS at x=b(the largest value supposedly) is 0 and RHS is b-a which is positive. This makes sure that this equation have real solutions. 
