$\bar{x}\int_{0}^{\bar{x}}f(x)\,dx>2\int_{0}^{\bar{x}}xf(x)\,dx$ always holds for positive and increasing $f(x)$? I come across one question that whether $\bar{x}\int_{0}^{\bar{x}}f(x)\,dx>2\int_{0}^{\bar{x}}xf(x)\,dx$ always holds. In this inequality, $f(x)$ is a strictly increasing and positive smooth function on $(0,\infty)$ and $\bar{x}\in (0,\infty)$. Is this inequality always right ? Can anyone help to prove it or find one counter-example ? Thank you. 
 A: It's the other way around: there holds 
$$x\int^x_0f(t)dt\leq 2\int_0^x tf(t)dt$$
if $f$ is increasing and positive. 
To see this, consider the function:
$$F(x)=2\int_0^x tf(t)dt-x\int^x_0f(t)dt.$$
Then 
$$\tag{1}F'(x)=2xf(x)-\int_0^xf(t)dt-xf(x)=xf(x)-\int_0^xf(t)dt.$$
Since $f$ is strictly increasing, we have $f(x)>f(t)$ for all $t\in(0,x)$. This implies that 
$$\tag{2}\int_0^xf(t)dt<f(x)\int_0^xdt=xf(x).$$
Combining $(1)$ and $(2)$, we have $F'(x)>0$. This implies that 
$$F(x)\geq \lim_{x\to 0^+}F(x)=0.$$ 
A: First, let $x > 0$ and we write
$$ I := 2 \int_{0}^{x} t f(t) \, dt - x \int_{0}^{x} f(t) \, dt = \int_{0}^{x} (2t-x)f(t) \, dt $$
Applying the substitution $t \mapsto x - t$, we also find that
$$ I := -\int_{0}^{x} (2t-x)f(x-t) \, dt. $$
Thus combining two computation, we have
$$ 2I = \int_{0}^{x} (2t-x)(f(t) - f(x-t)) \, dt. $$
But since $f$ is strictly increasing, it follows that $f(t) - f(x-t) > 0$ if and only if $t > x/2$. This shows that
$$ (2t-x)(f(t) - f(x-t)) \geq 0 \qquad \text{for } t \in (0, x) $$
and the equality holds if and only if $t = x/2$. Therefore $I > 0$ and hence
$$ 2 \int_{0}^{x} t f(t) \, dt > x \int_{0}^{x} f(t) \, dt. $$
Notice that in this proof, neither positivity nor smoothness is used.
