The problem and my proof in question:
If $a, b, c > 0$, show that it is not possible for each of the polynomials $P(x) = ax^2 + bx + c, Q(x) = cx^2 + ax + b, R(x) = bx^2 + cx + a$ to have two real roots.
Proof: WLOG let $a\geq b\geq c>0$. Then it follows that $c^2-4ab<0$. Thus, $Q(x)$ does not have 2 real roots.
My question is how do we justify saying WLOG in this situation. The reasoning I have built up so far seems incomplete. Here is my reasoning so far:
Clearly if we consider the two polynomials $R(x)=bx^2 + cx + a$ and $ax^2 + cx + b$ we see that the determinant, $c^2-4ab$ is the same for both. The same holds for $Q(x)$ and $P(x)$. So we have all $3!=6$ "coefficient permutations" for $a,b,$ and $c$. Therefore, we can justify substituting $a$ with $b$ or $c$ and vice verse and the problem will remain unchanged. Is there a better way of saying this? Am I wrong?
Extra: There is also an alternative proof to this problem:
Assume all given polynomials have two real roots. Then, $c^2>4ab, b^2>4ac, a^2>4bc$. Which implies the immpossible, namely that $a^2b^2c^2>64a^2b^2c^2$.
The WOLOG is not correct. Because of the cyclic nature of the problem you can relabel $a,b,c$ to make $a$ the greatest. Once you do that, there is no restriction on whether $b$ or $c$ is greater. I would say if $a \ge b \ge c \gt 0,$ then $c^2-4ab \lt 0$ and $Q(x)$ does not have real roots. If $a \ge c \ge b \gt 0,$ then $b^2-4ac \lt 0$ and $P(x)$ does not have real roots.
Alternately, you can make the argument SangchulLee makes in the comments to justify the reverse cyclic order. It is a good one, but I think it should be stated.