Write $P(x) = (x + a)(x + b)$ with $a,b > 1$ and $|a - b| < 2$. We are asked to prove that $P(P(x)) = ((x + a)(x + b) + a)((x + a)(x + b) + b)$ is positive for all real $x$. It suffices to prove that $(x + a)(x + b) + \min(a,b) > 0$ for any $x \in \mathbb{R}$. WLOG $a > b$. \begin{align*} (x + a)(x + b) + b &= \left( x + \frac{a + b}{2} \right)^2 - \frac{(a - b)^2}{4} + b \\ &\ge b - \frac{(a - b)^2}{4} \\ &> b - 1 \\ &> 0 \end{align*}

WLOG let $a$ be the greater root of the polynomial and the smaller root be $b$. Then $0 < a - b < 2 \iff 0 < r^2 - 4s < 4$. Clearly for since $P(x)$ is a monic quadratic polynomial, $P(x) > 0 \iff x > a$ or $x < b$. By the Trivial Inequality, the minimum value of $P(x)$ is as follows: \[ P(x) = \left (x + \frac{r}{2} \right)^2 + \frac{4s - r^2}{4} \geq \frac{4s - r^2}{4}. \]However $a < -1 < \frac{4s - r^2}{4} \leq P(x)$, so $P(P(x)) > 0$.

Interesting and good problem