By the Cauchy inequality,
$$(x+1)(y+1) \le \frac{((x+1) + (y+1))^2}{4} = \frac{(x+y+2)^2}{4}.$$
So the required inequality
$$\frac{1}{x+y+1} - \frac{1}{(x+1)(y+1)} < \frac{1}{11}, \quad \text{if } x > 0 \text{ and } y > 0, \quad (1)$$
is a consequence of the inequality
$$\frac{1}{x+y+1} - \frac{4}{(x+y+2)^2} < \frac{1}{11}, \quad \text{if } x > 0 \text{ and } y > 0. \quad (2)$$
Putting $x+y+1 = t$ ($t > 1$, since $x$ and $y$ are positive) into (2), we obtain the inequality
$$\frac{1}{t} - \frac{4}{(t+1)^2} < \frac{1}{11}, \quad \text{if } t > 1 \quad (3)$$
which is equivalent to (2). This inequality can be represented as
$$t^3 - 9t^2 + 23t - 11 > 0, \quad \text{if } t > 1. \quad (4)$$
Inequality (1) will be proved once we prove (4). We rearrange (4) as
$$t^3 - 9t^2 + 23t - 11 = (t-3)^3 - 4t + 16 = (t-3)^3 - 4(t-3) + 4 = (t-1)(t-3)(t-5) + 4.$$
Therefore, (4) is equivalent to the inequality
$$(t-1)(t-3)(t-5) + 4 > 0, \quad \text{if } t > 1. \quad (5)$$
We see that for $t > 1$ the polynomial $(t-1)(t-3)(t-5) \le 0$ only if $t \in [3; 5]$. But in this case $0 < t - 1 \le 4$ and $(t-3)(t-5) \ge -1$, therefore, $(t-1)(t-3)(t-5) \ge -4$ for $t \in [3; 5]$. Thus (5) is proved, which implies that (1) holds.