Consider the following expressions $$ \begin{align} x_1&\equiv \alpha^2 p+(1-\alpha)^2(1-p), \\ x_2&\equiv \alpha(1-\alpha) ,\\ x_3&\equiv (1-\alpha)^2 p+\alpha^2(1-p),\\ \end{align} $$ where $\alpha\in (1/2,1]$ and $p\in [0,1]$. Is it possible to establish an ordering among $x_1, x_2, x_3$? It is assumed that $x_1\in (0,1)$, $x_2\in (0,1)$, $x_3\in (0,1)$.
So far, I have tried to answer this question by attempting to solve the inequalities $$ x_1\geq (\text{ or, }\leq \text{ }) x_2\\ x_1\geq (\text{ or, }\leq \text{ }) x_3\\ x_2\geq (\text{ or, }\leq \text{ }) x_3\\ $$ but I did not go far because I end up with complicated expressions. I'm wondering whether there may be a smarter way to solve this exercise. Could you advise?
I think the best way to approach this is to think of $\alpha$ as fixed and $p$ as varying. Since $\frac12<\alpha\leq1$, we know that $\alpha>1-\alpha$, and consequently that $$(1-\alpha)^2<\alpha(1-\alpha)<\alpha^2$$
The middle term in this inequality is, of course, $x_2$. Now, when $p=0$, we have $x_1=(1-\alpha)^2,\ x_3=\alpha^2$ so that $$x_1<x_2<x_3.$$ When $p=1$, we have $x_1=\alpha^2,\ x_3=(1-\alpha)^2$ so that $$x_3<x_2<x_1.$$
Therefore, no order can be established among the three expressions.