Given 6 different real numbers $x_1,x_2,x_3,y_1,y_2,y_3$, Which satisfy the following conditions
$$ x_1 < y_1, x_2 < y_2, x_3 < y_3
$$
$$ x_1+y_1= x_2 + y_2= x_3 + y_3 $$
$$ x_1y_1+x_3 < y_3 = 2x_2y_2 > 0 $$
Which one of the following must be true?
A) $2x_2 < x_1+x_3$ B) $2x_2> x_1+x_3$ C) $x_2^2<x_1x_3$ D)$ x_2^2>x_1x_3$ E) $x_1^2>x_2x_3.$
I've been trying starting from some of the options and go backwards to see if I can get to a contradiction or something but I've never been able to get rid of all of the y's in the equations or inequalities when doing substitutions. I've also tried going from some of the conditions to any of the answer options but I arrive to the same problem.
I'd really appreciate your help. I'm preparing for an admission test and this is one of the past tests questions I haven't been able to solve.
Best Answer
In fact, none of these are necessarily true.
Consider the following assignments:
\begin{align} \tag{I} x_1 = \frac{1725}{4096} ,& y_1 = \frac{4547}{4096} ,& x_2 = \frac{49-\sqrt{353}}{64} ,& y_2 = \frac{49 + \sqrt{353}}{64} ,& x_3 = \frac{17}{32} ,& y_3 = 1\\ \tag{II} x_1 = \frac{-7}{4} ,& y_1 = \frac{-5}{4} ,& x_2 = \frac{-3-\sqrt{7}}{2} ,& y_2 = \frac{-3 + \sqrt{7}}{2} ,& x_3 = -4 ,& y_3 = 1 \end{align}
Both assignments satisfy the three conditions you listed. Moreover, assignment (I) satisfies both $2x_2 < x_1 + x_3$ and $x_2^2 < x_1x_3$, while assignment (II) satisfies both $2x_2 > x_1 + x_3$ and $x_2^2 > x_1x_3$.