Show $\displaystyle\int_0^af(x)g(x)dx\ge\int_0^af(a-x)g(x)dx$ 
Assume $f$ and $g$ are monotonically increasing on $[0,a]$, Show that
$$\displaystyle\int_0^af(x)g(x)dx\ge\int_0^af(a-x)g(x)dx$$

If I differentiate both sides w.r. to $a$ then;
$f(a)g(a)\ge f(0)g(a)$ and then integrate again gives the inequality ?
 A: HINT: Use the fact that $$\left(f(x)-f(a-x)\right)\left(g(x)-g(a-x)\right) \geq 0.$$ 
A: Consider
$$
\begin{align}
&\int_0^a[f(x)-f(a-x)]g(x)\,\mathrm{d}x\\
&=\int_{-a/2}^{a/2}[f(a/2+x)-f(a/2-x)]g(a/2+x)\,\mathrm{d}x\tag{1}\\
&=\frac12\int_{-a/2}^{a/2}[f(a/2+x)-f(a/2-x)][g(a/2+x)-g(a/2-x)]\,\mathrm{d}x\tag{2}\\
&=\int_0^{a/2}[f(a/2+x)-f(a/2-x)][g(a/2+x)-g(a/2-x)]\,\mathrm{d}x\tag{3}\\[9pt]
&\gt0\tag{4}
\end{align}
$$
Explanation:
$(1)$: substitute $x\mapsto a/2+x$
$(2)$: since $f(a/2+x)-f(a/2-x)$ is odd, we can replace $g(a/2+x)$ by its odd part
$(3)$: the product of two odd functions is even, so half the integral is over the positive domain
$(4)$: both $f$ and $g$ are monotonically increasing, so both factors in the last integrand are positive
A: Suppose that $$\int_{0}^{a}f(x)g(x)dx<\int_{0}^{a}f(a-x)g(x)dx.$$
Hence, the derivative with respect ot $a$ gives:
$$\left(\int_{0}^{a}f(x)g(x)dx\right)'<\left(\int_{0}^{a}f(a-x)g(x)dx\right)'.$$
$$\Leftrightarrow$$
$$f(a)g(a)<f(0)g(a).$$


*

*If $g(a)\leq0$: I do not know what to do.

*If $g(a)> 0$: then $f(a)<f(0)$ which is absurd since $f$ is monotonically increasing on $[0, a]$


Therefore:
$$\int_{0}^{a}f(x)g(x)dx\geq\int_{0}^{a}f(a-x)g(x)dx.$$
