1. We start by considering the given inequality:
$$\sum_{i=1}^n \frac{a_ib_i - b_i^2}{a_i + b_i} \leq \frac{\sum_{i=1}^n a_i \cdot \sum_{i=1}^n b_i - \left( \sum_{i=1}^n b_i \right)^2}{\sum_{i=1}^n (a_i + b_i)}$$

2. First, we use the T2 Lemma (a form of the Cauchy-Schwarz inequality) to handle the term involving $b_i^2$:
$$\sum_{i=1}^{n} \frac{-b_{i}^{2}}{a_{i}+b_{i}} \leq \frac{-\left( \sum_{i=1}^{n} b_{i} \right)^2}{\sum_{i=1}^{n} (a_{i} + b_{i})} \quad (\star)$$

3. For brevity, let:
$$\theta = \sum_{i=1}^{n} a_{i} \quad \text{and} \quad \Omega = \sum_{i=1}^{n} b_{i}$$

4. Next, we consider the term involving $a_i b_i$:
$$\sum_{i=1}^{n} \frac{(a_{i} + b_{i})^2}{4(a_{i} + b_{i})} - \sum_{i=1}^{n} \frac{a_{i} b_{i}}{a_{i} + b_{i}} = \sum_{i=1}^{n} \frac{(a_{i} - b_{i})^2}{4}$$

5. Applying the T2 Lemma again, we get:
$$\sum_{i=1}^{n} \frac{(a_{i} - b_{i})^2}{4} \geq \frac{(\theta - \Omega)^2}{\theta + \Omega}$$

6. Rearranging the above inequality, we obtain:
$$\sum_{i=1}^{n} \frac{a_{i} b_{i}}{a_{i} + b_{i}} \leq \frac{\theta + \Omega}{4} - \frac{(\theta - \Omega)^2}{\theta + \Omega}$$

7. Simplifying the right-hand side:
$$\frac{\theta + \Omega}{4} - \frac{(\theta - \Omega)^2}{\theta + \Omega} = \frac{\theta \cdot \Omega}{\theta + \Omega}$$

8. Thus, we have:
$$\sum_{i=1}^{n} \frac{a_{i} b_{i}}{a_{i} + b_{i}} \leq \frac{\sum_{i=1}^{n} a_{i} \cdot \sum_{i=1}^{n} b_{i}}{\sum_{i=1}^{n} (a_{i} + b_{i})} \quad (\dagger)$$

9. Adding the inequalities from $(\star)$ and $(\dagger)$, we get:
$$\sum_{i=1}^n \frac{a_i b_i - b_i^2}{a_i + b_i} \leq \frac{\sum_{i=1}^n a_i \cdot \sum_{i=1}^n b_i - \left( \sum_{i=1}^n b_i \right)^2}{\sum_{i=1}^n (a_i + b_i)}$$

10. The equality holds when $a_i = b_i$ for all $1 \leq i \leq n$.

$\blacksquare$