Let $X_{1},X_{2},\ldots,X_{n}$ be independently drawn from a distribution $F$ and let $Y_{k}^{(n)}$ be the $k$-th order statistic (Convention: $Y_{1}^{(n)}>Y_{2}^{(n)}>\cdots>Y_{n}^{(n)}$). What is the probability distribution (or the density if this is simpler to write down?) of the maximum of $Y_{1}^{(n)}$ and $Y_{2}^{(n)}+a$ with $a$ some positive constant.
So how is $\max\{Y_1^{(n)},Y_2^{(n)}+a\}$ distributed? I know that the density of $Y_2^{(n)}$ conditional on $Y_1^{(n)}=c$ is given by \begin{equation} f_2^{(n)}\Big(x \,\big\vert\, Y_1^{(n)}=c\Big)=\frac{f_{1,2}^{(n)}\big(c,x\big)}{f_1^{(n)}\big(c\big)}, \end{equation}where $f_{1,2}^{(n)}\big(x,y\big)$ is given by\begin{equation} f_{1,2}^{(n)}\big(x,y\big)=\begin{cases} n(n-1)f(x)f(y)F(y)^{n-2} & \text{ for }x\geq y\\ 0& \text{ else.} \end{cases} \end{equation}But I'm not sure on how to proceed from here.
$\mathbb P(\max(Y_1, Y_2 + a)\le x) = \mathbb P(Y_1 \le x, \; Y_2 \le x - a)$ is the probability that there is at most one $X_i$ in $(x-a,x]$ and none in $(x, \infty)$. The probability that all are in $(-\infty, x-a]$ is $F(x-a)^n$ (where $F$ is the cdf of your distribution), while the probability that exactly one is in $(x-a,x]$ and the others in $(-\infty, x-a]$ is $n F(x-a)^{n-1} (F(x) - F(x-a))$. Thus the total is $$nF(x-a)^{n-1} F(x) - (n-1) F(x-a)^n$$