\begin{array}{ll} \text{maximize} & \frac{x_1x_2}{y}\\ \text{subject to} & b_1+b_2\geq x_1+x_2\\&b_2\geq x_2\\&b_1\geq y\\&x_1\geq y\geq x_2\end{array}
where $x_i,y,b_i\in \mathbb{R}^+$, $b_1$,$b_2$ are given s.t $b_1\geq b_2$.
My attempt: Let $x_2+k=b_2$, where $b_2>k\geq0$, then \begin{array}{ll} \text{maximize} & \frac{(b_1+k)(b_2-k)}{y}\\ \text{subject to} &b_1\geq y\geq b_2-k\end{array} $\max (b_1+k)(b_2-k)=b_1b_2$ when $k=0$, but $\min y$ is achieved when $k$ is not zero but as close as possible to $b_2$. I can't figure out how to find compromise here.