Now $1\leq a_1<a_2<\cdots <a_n$ So $a_m\leq a_n+(m-n)$ where $1\leq m\leq n$ and $n\leq a_n$ So $\sum^{n-1}_{i=1}a_i\leq (n-1)a_n-\sum^{n-1}_{i=1} i =(n-1)a_n-\frac{n(n-1)}{2}$ So $\frac{2n-4}{n-1}\sum^{n-1}_{i=1}a_i\leq 2(n-2)a_n-n(n-2)=(n-2)(2a_n-n)$ $n\leq a_n$ so $a_n-n+2>a_n-n\geq 0$ so $(a_n-n+2)(a_n-n)\geq 0$ So $a_n^2-2a_n\geq (n-2)(2a_n-n)$ So $a_n^2\geq \frac{2n-4}{n-1}\sum^{n-1}_{i=1}a_i +2a_n$ So $MAX(\lambda)\geq\frac{2n-4}{n-1}$ Consider $a_m=m$ where $1\leq m\leq n$ Then $n(n-2)=n^2-2n\geq \lambda \cdot \frac{n(n-1)}{2}$ Then $\lambda \leq\frac{2n-4}{n-1}$ So $MAX(\lambda)\leq\frac{2n-4}{n-1}$ Hence $\frac{2n-4}{n-1}\leq MAX(\lambda)\leq\frac{2n-4}{n-1}$ Hence $MAX(\lambda)=\frac{2n-4}{n-1}$