Let $\int_{- \infty}^{\infty} f(x) dx =1$. Then show that $ \int_{- \infty}^{\infty} \frac{1}{1+ f(x)} dx = \infty.$
Solution 1:
$$\int_{-c}^c \frac{1}{1+f(x)}\, dx = \int_{-c}^c \frac{1+f(x)}{1+f(x)} \, dx - \int_{-c}^c \frac{f(x)}{1+f(x)} \, dx \\ = 2c - \int_{-c}^c \frac{f(x)}{1+f(x)} \, dx \\ \geqslant 2c - \int_{-c}^c f(x) \, dx$$
Solution 2:
The set $E=\{x\,:\,f(x)>1\}$ has finite measure since otherwise $\int_{\mathbb{R}} f\ge \int_E f \ge |E| = \infty$. So $\mathbb{R}\backslash E$ has infinite measure and thus we get
$$\int_{-\infty}^\infty \frac1{1+f(x)} dx \ge \int_{\mathbb{R}\backslash E} \frac{1}{2} dx = \infty$$
because $f(x)\le 1$ for $x\in \mathbb{R}\backslash E$.