Since $\sum abc\le \sum (k+\alpha)a\sum(k+\beta)b\sum(k+\gamma)c\le\alpha\beta\gamma<1$, $\lambda\ge\frac{\sum kabc}{1-\sum abc}$. Now our goal is to find maximum of $I= \frac{\sum kabc}{1-\sum abc}$. $I=\alpha\beta\gamma\frac{\sum kabc}{\alpha\beta\gamma-\alpha\beta\gamma\sum abc} \le\frac{\alpha\beta\gamma\sum kabc}{\sum (k+\alpha)a\sum(k+\beta)b\sum(k+\gamma)c-\alpha\beta\gamma\sum abc}\le\frac{\alpha\beta\gamma\sum kabc}{\sum [(k+\alpha)(k+\beta)(k+\gamma)-\alpha\beta\gamma] abc}\le \alpha\beta\gamma\max_{1\le k\le n}\frac{k}{[(k+\alpha)(k+\beta)(k+\gamma)-\alpha\beta\gamma]}.$ The last step uses fact $\frac{x+y}{a+b}\le\max(\frac xa,\frac yb)$. Now we see $I\le\frac{\alpha\beta\gamma}{(1+\alpha)(1+\beta)(1+\gamma)-\alpha\beta\gamma}$. It is attainable by letting $a_1=\frac{\alpha}{1+\alpha}$, $b_1=\frac{\beta}{1+\beta}$, $c_1=\frac{\gamma}{1+\gamma}$ and others zero.