Note that it suffices to prove that .

WLOG let , and let , . It is well known that lies on and lies on and that , are both isosceles. (The proof of this basically comes down to showing that and similarly for the triangles associated with .) Thus . Next, we know that by the definition of reflection and , so . Similarly, , so and . Therefore by the Extended Law of Sines, we have
where denotes the circumradius of .

Finally, let and be the feet of the projections from and to and respectively. We know that by our work above, and in addition we know that . Therefore by Pythag or HL or whatever we have . Thus since , we have that , as desired.

\[\left\{\begin{array}{c c c} \left(a+b\right)\left(b+c\right)\left(c+a\right)&=&abc,\\ \left(a^3+b^3\right)\left(b^3+c^3\right)\left(c^3+a^3\right)&=&a^3b^3c^3.\end{array}\right\,
\]

Suppose that . Dividing both sides of the first equation by gives
and a similar result can be found by dividing from the second equation. Let , , . Then it suffices to prove that no exist such that
Once again, suppose such exist. Then
If are all positive, then by AM-GM. Otherwise, WLOG let . Then and similarly for , so the LHS is greater than or equal to Thus, the only equality case occurs when , but then , contradiction. Hence such do not exist, and the only way this can happen from our original substitutions occurs when at least one of equals zero, which is what we wanted.

Note that . Therefore
Since and , in order to show that this is composite we must show that is not possible. Suppose it is possible. Then squaring yields
Using the fact that again gives
However, since we have a contradiction. Therefore and is composite, as desired.

From we know that . Substituting this into the second equation gives
Similarly, and . Now it suffices to maximize the square of our product, which is
Expansion gives

\[\begin{align*}(1-xy)(1-yz)(1-zx)&=1-xy-yz-zx+x^2yz+xy^2z+xyz^2-(xyz)^2\\&=1-(-3)+0-(xyz)^2=4-(xyz)^2.\end{align*}\]

By the Trivial Inequality, , so
which means that . Equality holds when, say, .

Let the parallel to through intersect at and at . I claim that is the midline of w.r.t , which implies the result.

Note that is a trapezoid with bases and , so (). Therefore by the definition of point we have
so is the midpoint of . Similarly, is the midpoint of . Therefore is a midline of as desired.

Since is a right triangle with hypotense , , so it suffices to show that . Let denote the projection of to the segment . Remark that since and , we have . Furthermore, since is tangent to and since is cyclic, we have
Therefore and
as desired.

This was the original solution I found; I'm aware that this solution can be shortened by considering as opposed to . The idea is the same though.