We will use the following properties:1. For any positive number a and b, we have a^2+b^2 \geq 2ab 2. For any non-negative numbers x and y, we have x+y \geq 2\sqrt{xy}We can rewrite the inequality as follows:\frac{1-(\lambda-(m+1)) \cdot \frac{m!}{\lambda^{m+1}} \cdot \sum_{j=0}^{m} \frac{\lambda^j}{j!}}{\frac{\lambda}{(\lambda-(m+1))^2}} \leq 1 Using property 1 we obtain:(1-(\lambda-(m+1)) \cdot \frac{m!}{\lambda^{m+1}} \cdot \sum_{j=0}^{m} \frac{\lambda^j}{j!})^2 \leq (\frac{\lambda}{(\lambda-(m+1))^2})^2 Expanding the left hand side of this inequality gives us:1-2(\lambda-(m+1)) \cdot \frac{m!}{\lambda^{m+1}} \cdot \sum_{j=0}^{m} \frac{\lambda^j}{j!}+(\lambda-(m+1))^2 \cdot \frac{m!^2}{\lambda^{2(m+1)}} \cdot \left(\sum_{j=0}^{m} \frac{\lambda^j}{j!}\right)^2 \leq \frac{\lambda^2}{(\lambda-(m+1))^4}Using property 2 for the term in the middle on the left hand side of the inequality gives us:1-2(\lambda-(m+1))^2 \cdot \frac{m!^2}{\lambda^{2(m+1)}} \cdot \sum_{j=0}^{m} \frac{\lambda^j}{j!} +(\lambda-(m+1))^2 \cdot \frac{m!^2}{\lambda^{2(m+1)}} \cdot \left(\sum_{j=0}^{m
