MHB Can This Complex Inequality Be Proven?

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    Inequality Proof
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Hallo, can someone help me to proof this inequality:

$$ 1-(\lambda-(m+1)) \cdot \frac{m!}{\lambda^{m+1}} \cdot \sum_{j=0}^{m} \frac{\lambda^j}{j!} \leq \frac{\lambda}{(\lambda-(m+1))^2} $$ under condition $$ m+1 < \lambda $$.
 
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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
 
I posted this question on math-stackexchange but apparently I asked something stupid and I was downvoted. I still don't have an answer to my question so I hope someone in here can help me or at least explain me why I am asking something stupid. I started studying Complex Analysis and came upon the following theorem which is a direct consequence of the Cauchy-Goursat theorem: Let ##f:D\to\mathbb{C}## be an anlytic function over a simply connected region ##D##. If ##a## and ##z## are part of...
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