<s>[INST] 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?
 [/INST] 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. 
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