TITLE: A game-theoretical question in a political economy model
QUESTION [7 upvotes]: My research question in a dynamic model of political competition boils down to the following conjecture. I am confident that it holds (all simulations work), but I have not been able to prove it yet. Let $c,\eta,\mu,i$ be parameters such that $0<c$, $1<\eta \leq 2$, and $0<i<\mu$. Define $r\in (i,\mu)$ and $l\in (-\mu,i)$ such that $$2(\mu-r)=c\eta (r-i)^{\eta-1}$$ and $$2(l+\mu)=c\eta (i-l)^{\eta-1}.$$
It is easy to see that $r$ and $l$ are uniquely defined within the given ranges. Then, I would like to prove that
$$r^2+c(r-i)^{\eta} \leq l^2+c(i-l)^{\eta}.$$ 
In this problem, $i$ is the status-quo policy, $c$ and $\eta$ determine the cost of moving policies, $\mu$ is the degree of polarization of political parties, and $r$ ($l$) is the policy that a right-wing (left-wing) candidate would choose if elected. The inequality that I want to prove then simply says that the utility that the median voter will derive if $r$ is elected is at least as much as the one he/she will derive if $l$ is elected. We can assume that $r$ is the incumbent. If $\eta=2$, the result is straightforward. Together with my coauthors, I have already spent some time (in vain) trying to prove the claim. I have the impression that there has to be some easy argument, but in all proof strategies I have tried expressions turn extremely long and cumbersome immediately. By simulations, I also know that if $\eta>2$ or $i<0$ the result does not hold, so $1<\eta\leq2$ and $0<i<\mu$ should be used in the proof in some step. Any help or hint would be greatly appreciated. Thank you for your time!

REPLY [5 votes]: The following seems to work:
Change variables to simplify:  $x := r-i$, $y := i-\ell$, $p := \eta - 1$, $a := \mu - i$, $b := \mu + i$.  Your two equations defining $r$ and $\ell$ can then be solved for $a$ and $b$ and the result used to express the desired inequality in terms of $x$, $y$, $c$, and $p$ (eliminating $a$ and $b$).  Then the constraints reduce to $0 < c$ and $0 < p \leq 1$ and $0<x<y$, and the inequality to be proved becomes
$$
 f(x,y) := \left[\left(\frac{x+y}{2}+\frac{c}{4}(p+1)(x^p-y^p)\right)^2+cy^{p+1}\right]-\left[\left(\frac{x+y}{2}+\frac{c}{4}(p+1)(y^p-x^p)\right)^2+cx^{p+1}\right] \geq 0.
$$
You've already handled the $p=1$ case, so assume $p<1$.  Clearly, $f(x,x)=0$.  By differentiating with respect to the second slot of $f$, you can verify that the partial derivative $f_2(x,x) \geq 0$ and that the second-order partial derivative $f_{22}(x,y) \geq 0$ for every $y \geq x$.  Thus, $f_2(x,y) \geq 0$ for every $y \geq x$, and consequently $f(x,y) \geq 0$ for every $y \geq x$.
(Details of the calculations available on request.)