The following is taken from Baye, Shin (1999)
Consider a contest over a prize valued at 1 with symmetric players $1$ and $2$ who exert a level of effort $x_1$ and $x_2$ respectively. Effort cannot exceed $2/3$. Profit ($\pi$) of player 1 is
$$\pi_1=\frac{x_1-\frac{x_1x_2}{2}}{x_1+x_2-x_1x_2}-x_1$$
The simultaneous-move equilibrium is (denoted with superscript $*$)
$$x_1^*=x_2^*=1-\frac{\sqrt{2}}{2}$$
The profit for each is
$$\pi_1^*(x_1^*,x_2^*)=\pi_2^*(x_1^*,x_2^*)=\frac{\sqrt{2}-1}{2}$$
And we have that the best response of player $2$
$$R_2(x_1)=\frac{2x_1-\sqrt{(-2x_1^2+4x_1)}}{2(x_1-1)}$$
However, suppose that player $1$ 'moves' before player $2$. Then, player $1$ would deviate (upwards) from $x_1^*$, presumably, because player $2$ would reduce it's level of effort, and it would receive more of the prize. How does one show this?
In the article, Baye and Shin consider the new Stackelberg equilibrium $\pi_1^s$ where they show that $\pi_1^s(x_1^*+\epsilon)>0$ for some $\epsilon>0$. Specifically
$$\pi_1^s(x_1^*+\epsilon)-\pi_1^*(x_1^*+x_2^*)=\frac{\sqrt{1+2\epsilon\sqrt{2}-2\epsilon^2}-1-\epsilon\sqrt{2}+2\epsilon^2}{\sqrt{2}-2\epsilon}>0$$
I've tried replicating their result by seeing how $x_2$ changes with $x_1+\epsilon$ and substituting $x_1+\epsilon$ and the new value of $x_2^s$ into $1$'s profit function, but I fail to reach their result.
