Show $\displaystyle\int_0^af(x)g(x)dx\ge\int_0^af(a-x)g(x)dx$

Question

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 ?

Answer 1

HINT: Use the fact that $$\left(f(x)-f(a-x)\right)\left(g(x)-g(a-x)\right) \geq 0.$$

Answer 2

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

Answer 3

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