Proving an inequality about the product of integrals

$$ \langle f, g \rangle = \int_{-1}^1 f(x) g(x) \frac{\exp(a + bx)}{(1 + \exp(a + bx))^2}dx $$ is an inner product on the space $C([-1, 1])$ of continuous functions on the interval $[-1, 1]$, therefore the Cauchy-Schwarz inequality can be applied: $$ |\langle f, g \rangle |^2 \le \langle f, f \rangle \cdot \langle g, g \rangle \, . $$ Choosing $f(x) =1 $ and $g(x) = x$ gives the desired inequality. Strict inequality holds because the functions are not a constant multiple of each other.

Or course one can also apply the “usual” Cauchy-Schwarz inequality $$ \left(\int_a^b F(x) G(x) dx \right)^2 \le \int_a^b F(x)^2 dx \cdot \int_a^b G(x)^2 dx $$ to $$ F(x) =\frac{\sqrt{\exp(a + bx)}}{1 + \exp(a + bx)} \, , \, G(x) =x \cdot \frac{\sqrt{\exp(a + bx)}}{1 + \exp(a + bx)} \, . $$

If $u$ is a function such that $u'(x) =\frac {a+bx}{(1+exp(a+bx))^2}$, and $l = u^{-1}(-1), r = u^{-1}(1)$, then your inequality becomes $\left(\int_l^rdu\right)\left(\int_l^rx^2du\right)>\left(\int_l^rxdu\right)^2$.

So if you can show that $u$ and $u^{-1}$ exist, you can apply the Cauchy-Schwarz inequality.