The question is if $x,y,z$ are independent $x\sim\exp(\lambda), y\sim\exp(\mu), z\sim\exp(\gamma)$ and define $u=\min(x,y), v=\min(y,z)$ what is the probability $p(U>u,V>v)$. Consider the cases $u>v,v>u$ separately.
Is the answer $p(U>u,V>v):$
For $u>v \implies \exp[-(\mu+\lambda)u]\exp[-(\gamma)v]$
For $v>u \implies \exp[-(\mu+\gamma)v]\exp[-(\lambda)u]$
Then if we differentiate $p(U>u,V>v)$ this give the density distribution as it is required to prove that as attached in equation where $g(u,v)$ is the density distribution but the integration results in 1 not as he wanted so where is the error!
$$\begin{align} \text{(a)}& \text{Find }P(U>u,V>v) \text{ for } u,v>0.\text{ Consider the cases }u>v\text{ and }u<v\text{ separately.}\\ \text{(b)}&\text{Differentiate (a) with respect to }u \text{ and }v,\text{ obtaining a "joint density" function,}\\ &g(u,v) \text{ say}.\\ \text{(c)}&\text{Show that }\int_0^\infty\int_0^\infty g(u,v)\text{ }du\text{ }dv\text{ }= (\gamma+\lambda)/(\gamma + \lambda + \mu).\text{ Why isn't this integral}\\ &\text{equal to 1?} \end{align}$$