Prove that $\int_{-a}^{b} x^2f(x)dx\leq ab\int_{-a}^{b}f(x)dx$. 
Let $a,b>0$,$f\in C[-a,b]$,$f>0$,$$\int_{-a}^{b} xf(x)dx=0,$$then prove that $$\int_{-a}^{b} x^2f(x)dx\leq ab\int_{-a}^{b}f(x)dx,$$

I have tried with the $$\int_{-a}^{b} x^2f(x)dx=\varepsilon^2\int_{-a}^{b}f(x)dx,$$but it doesn't work because I can't compare the value of the ab with the $\varepsilon$,And I don't how to use the condition"f is continuous" by this method.
 A: For any $x\in[-a,b]$, we have
$$ab-x^2+(b-a)x=(x+a)(b-x)\geq 0\,.$$
For any continuous function $f:[-a,b]\to\mathbb{R}$ such that $f(x)\geq 0$ for all $x\in[-a,b]$, we then have
$$\big(ab-x^2+(b-a)x\big)\,f(x)\geq 0$$
for every $x\in[-a,b]$.  Therefore,
$$\int_{-a}^b\,\big(ab-x^2+(b-a)x\big)\,f(x)\,\text{d}x\geq 0\,.$$
This means
$$ab\,\int_{-a}^b\,f(x)\,\text{d}x-\int_{-a}^b\,x^2\,f(x)\,\text{d}x+(b-a)\,\int_{-a}^b\,x\,f(x)\,\text{d}x\geq 0\,.$$
If $\displaystyle\int_{-a}^b\,x\,f(x)\,\text{d}x=0$, then it follows that
$$ab\,\int_{-a}^b\,f(x)\,\text{d}x-\int_{-a}^b\,x^2\,f(x)\,\text{d}x\geq 0\,,$$
which is equivalent to the inequality to be proven.  (The equality holds if and only if $f\equiv 0$.)
Remark.  The same claim holds if $f:[-a,b]\to\mathbb{R}_{\geq 0}$ is Riemann- or Lebesgue-integrable, and satisfies $\displaystyle\int_{-a}^b\,x\,f(x)\,\text{d}x=0$.  The equality holds if and only if $f=0$ almost everywhere.  If $f$ is a distribution, however, then the equality case is when $f(x)$ is a scalar multiple of $b\,\delta(x+a)+a\,\delta(x-b)$, where $\delta$ is the Dirac $\delta$-distribution.
