I assume that the reader understands by Rearrangement Inequality that if $a_i$ and $b_i$ are reals, and $a_1 ≤ a_2 ≤ ... ≤ a_n $ and $b_1 ≤ b_2 ≤ ...≤ b_n$ then
$\Sigma_{i=0}^{i=n} a_i × b_i$ ≥ $\Sigma_{i=0}^{i=n} a_i × p_i$ ≥ $\Sigma_{i=0}^{i=n} a_i × b_{n-i}$
Where $p_1,p_2...$ are any permutations of $b_1, b_2...$
Now Statement 1: Consider 3 positive numbers $a, b$, and $c$.
WLOG, assume $a ≥ b ≥ c > 0 $ Now, $1/c ≥ 1/b ≥ 1/a$
Applying Rearrangement Inequality, we get
$a/c + b/b + c/a ≥ a/b + b/a + c/c$
Or, $a/c + c/a ≥ a/b + b/a$
Now, the dilemma, if we had considered WLOG $a≥c≥b$ THEN the inequality would have flipped but the result should be independent of the order of $b$ and $c$ as we assumed the order without loss of generality...
I couldn't find where's the flaw in this that leads to the dilemma?
If $a≥b≥c>0$ then $a/c+c/a ≥ a/b+b/a$.
If $a≥c≥b>0$ then $a/b+b/a ≥ a/c+c/a$.
No contradiction there...