Integral inequation In my statistics book Chebyshev's inequality is proven. In several steps this inequality is used:
$$ \int_a^{+\infty} \phi(x) f_X(x)dx \quad \geq \quad \phi(a) \int_a^{+\infty} f_X(x)dx $$
and also:
$$ \int_{-\infty}^{-a} \phi(x) f_X(x)dx \quad \geq \quad \phi(-a) \int_{-\infty}^{-a} f_X(x)dx $$
Here is $a \geq 0$, $\phi:\mathbb{R}\to\mathbb{R}^+$ a positive function, and $f_X$ a pdf.
Why this is valid?
 A: Since $\phi$ is increasing (this is important here) and $f_X$ is positive, we have for $a>0$ and $x\ge a$, $\phi(x)\ge\phi(a)$. Therefore,
$$
\begin{align}
\int_a^\infty\phi(x)f_X(x)\,\mathrm{d}x
&\ge\int_a^\infty\phi(a)f_X(x)\,\mathrm{d}x\\
&=\phi(a)\int_a^\infty f_X(x)\,\mathrm{d}x\\
\end{align}
$$
The other inequality is simply a change of variables.
A: The intuitive reason why $$ \int_{a}^{+\infty} \phi(x) f_X(x)dx \quad \geq \quad \phi(a) \int_{a}^{+\infty} f_X(x)dx $$ is that we are integrating $\phi(x)$ from $a$ to $+\infty$, hence if we evaluate $\phi(x)$ at the lower limit of that integral ($\phi(a)$ is $\phi(x)$ evaluated at the lower limit of the integral), this result will always be smaller than the integral itself. You can see this best by making a picture of a positive valued function and see in the picture what the integral represents and what the value of the function evaluated at the lower limit of the integral represents. Though this is not a formal proof, it might help your understanding.
