$f(x)$ is a continuous function on $[0,\infty)$, and it is decreasing on its domain.
Then how to prove $\int_0^x{x^2f(t)\,dt}\ge3\int_0^x{t^2f(t)\,dt}$ when $x\ge0$ ?
Maybe I can have $F(x)=\int_0^x{x^2f(t)\,dt}-3\int_0^x{t^2f(t)\,dt}$ and work on its derivative, but what's the derivative of $\int_0^x{x^2f(t)\,dt}$? Can Fundamental Theorem of Calculus be used here? Or are there other ways to prove that?
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Edit:(This solution only works if $f$ is differentiable and $f'g+g'f$ is continuos where g(t)=(t*x^2-t^3))
Tip: Wirte the Integral as $\int_0^x(x^2-3t^2)f(t)dt$ (if your inequality holds this has to be $\geq 0$). Then use Partial integration. Since $f'\leq 0$ the assertion follows.
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This inequality is about rearrangment. The following generalization is true. Assume that $g,h\ge 0$ are continuous and $\int_0^t g(s)ds \ge \int_0^t h(s)ds$ for all $t\le x$ and $\int_0^x g(s)ds = \int_0^x h(s)ds$, then it holds
$$
\int_0^x g(s)f(s)ds\ge \int_0^x h(s)f(s)ds.
$$ If $f'$ exists and is continuous, then the claim follows by integration by parts formula.
Proof: We may assume $x=1$. Since $f$ is continuous, we can approximate $f$ uniformly by a step function of the form $$ s_n(x) = \sum_{j=1}^n f(\frac{j}{n})1_{[\frac{j-1}{n},\frac{j}{n}]}(x). $$ Let $D(t) = \int_0^t (g(s)-h(s))ds$. Then, $$ \int_0^1 (g(s)-h(s))s_n(s)ds = \sum_{j=1}^n f(\frac{j}{n})(D(\frac{j}{n})-D(\frac{j-1}{n}))=\sum_{j=1}^n D(\frac{j}{n})(f(\frac{j}{n})-f(\frac{j+1}{n}))\ge 0. $$ By taking $n\to \infty$, we get $$ \int_0^1 g(s)f(s)ds\ge \int_0^1 h(s)f(s)ds. $$
Your function $F$ is simply $F(x) = x^2 \int_0^x f(t)\, dt - 3 \int_0^x t^2 f(t)\, dt$. Since $f$ is continuous, you can differentiate $F$ obtaining $$ F'(x) = 2x \int_0^x f(t)\, dt + x^2 f(x) - 3 x^2 f(x). $$ Since $f$ is decreasing, you have that $f(t) \geq f(x)$ for every $t\in[0,x]$, hence $$ \int_0^x f(t)\, dt \geq x\, f(x). $$ You then obtain $F'(x) \geq 0$ for every $x\geq 0$. Since $F(0) = 0$, you can conclude that $F(x) \geq 0$ for every $x\geq 0$.