Prove that $\int_a^bxf(x)dx\geq\frac{b+a}{2}\int_a^bf(x)dx$ Let $f:[a,b]\to\mathbb{R}$ be continuous and increasing, show that
$$\int_a^bxf(x)dx\geq\frac{b+a}{2}\int_a^bf(x)dx$$
I am thinking of using integration by parts. First let
$$F(x)=\int_a^xf(t)dt$$
Then
$$\int_a^bxf(x)dx=bF(b)-\int_a^bF(x)dx=b\int_a^bf(x)dx-\int_a^b\int_a^xf(t)dtdx$$
So far I only have these in mind. Any one have any hints?
 A: You want to show that 
$$
I = \int_a^bxf(x)dx - \frac{b+a}{2}\int_a^bf(x)dx = \int_a^b \bigl(x - \frac{b+a}{2} \bigr) f(x) dx 
$$
is $\ge 0$. First split the integral in two parts:
$$ 
I =  \int_a^{(a+b)/2} \bigl(x - \frac{b+a}{2} \bigr) f(x) dx + \int_{(a+b)/2}^b \bigl(x - \frac{b+a}{2} \bigr) f(x) dx
$$
Now substitute $x = a+b-y$ in the first integral:
$$
 I = \int_{(a+b)/2}^b \bigl(\frac{b+a}{2} - y\bigr) f(a+b-y) dy + \int_{(a+b)/2}^b \bigl(x - \frac{b+a}{2}\bigr) f(x) dx \\
= \int_{(a+b)/2}^b \bigl(x - \frac{b+a}{2}\bigr) \bigl( f(x) - f(a+b - x)\bigr) dx \ge 0
$$
because $x \ge a+b-x$ in $[(a+b)/2 , b]$ and $f$ is increasing.
A: Using Chebyshev'inequality, we already have $f(x)$ and $g(x)=x$ are monotone, so $$\frac{1}{b-a}\int_a^b xf(x) \geq \left(\frac{1}{b-a}\int_a^b f(x)\right)\left(\frac{1}{b-a}\int_a^b x\right).$$
Then one has the conclusion.
A: Since $f$ is increasing, we have
$$
\frac1{b-t}\int_t^b\,f(x)\,\mathrm{d}x
\ge\frac1{b-a}\int_a^b\,f(x)\,\mathrm{d}x\\
$$
for $a\le t\le b$. Therefore,
$$
\begin{align}
\int_a^b(x-a)\,f(x)\,\mathrm{d}x
&=\int_a^b\int_a^x\,f(x)\,\mathrm{d}t\,\mathrm{d}x\\
&=\int_a^b\int_t^b\,f(x)\,\mathrm{d}x\,\mathrm{d}t\\
&\ge\int_a^b\frac{b-t}{b-a}\int_a^b\,f(x)\,\mathrm{d}x\,\mathrm{d}t\\
&=\frac1{b-a}\int_a^b(b-t)\,\mathrm{d}t\int_a^b\,f(x)\,\mathrm{d}x\\
&=\frac{b-a}2\int_a^b\,f(x)\,\mathrm{d}x
\end{align}
$$
Adding $a\int_a^bf(x)\,\mathrm{d}x$ to both sides, we get
$$
\int_a^bx\,f(x)\,\mathrm{d}x\ge\frac{b+a}2\int_a^b\,f(x)\,\mathrm{d}x
$$
A: Note that $\int_{a}^{b} \left(x - \frac{a + b}{2}\right)dx = 0$. Since $f$ is continuous on a compact set it is bounded and it achieves minimum and since it is increasing the minimum is achieved at the lowest end of an interval and maximum at the highest end of an interval.
Let $I_{1} = \left[a, \frac{a + b}{2} \right]$ and $I_{2} = \left[\frac{a + b}{2}, b\right]$, and $g(x) = \left(x - \frac{a + b}{2} \right)$.
We have  $g(x) \leq 0, f(x) \leq f(\frac{a + b}{2})$ on $I_{1}$. So that $g(x)f(x) \geq f(\frac{a + b}{2}) g(x)$.
Similarly, $g(x) \geq 0$,  $f(x) \geq f(\frac{a + b}{2})$ on $I_{2}$. So that $g(x) f(x) \geq f(\frac{a + b}{2})g(x)$ on $I_{2}$. 
Now we have:
$\int_{a}^{b}\left(x - \frac{a + b}{2} \right) f(x) dx$
$=  \int_{a}^{\frac{a + b}{2}}\left(x - \frac{a + b}{2} \right) f(x) dx  +  \int_{\frac{a + b}{2}}^{b}\left(x - \frac{a + b}{2} \right) f(x) dx$
$\geq f(\frac{a + b}{2})\left[\int_{I_{1}}g(x)dx + \int_{I_{2}} g(x) dx \right]$
$ = f(\frac{a + b}{2} )\int_{a}^{b} g(x)dx = 0$, which implies the desired inequality. 
A: There is a simple solution for this problem. Let
$$ F(x)=\int_a^xf(t)dt-\frac{a+x}{2}\int_a^xf(t)dt. $$
Then
$$ F'(x)=xf(x)-\frac{1}{2}\int_a^xf(t)dt-\frac{a+x}{2}f(x)=\frac{1}{2}\int_a^x[f(x)-f(t)]dt. $$
Noting that $f(t)$ is increasing in $[a,b]$ and $a\le t\le x$, we have $f(x)\ge f(t)$ and hence $F'(x)\ge 0$ or $F(x)$ is increasing. So $F(b)\ge F(a)=0$ is the desired result.
A: Actually, you're almost there. Since $f$ is increasing, we have
$$ F(x) = \int\limits_{a}^{x}{f(t)\text{ d}t} \le \int\limits_{a}^{x}{f(x)\text{ d}t} = (x-a)f(x) $$
and hence
$$\int\limits_{a}^{b}{F(x)\text{ d}x}\le\int\limits_{a}^{b}{(x-a)f(x)\text{ d}x}.$$
Hence, we have
\begin{align*}\int\limits_{a}^{b}{xf(x)\text{ d}x} = b\int\limits_{a}^{b}{f(x)\text{ d}x} - \int\limits_{a}^{b}{F(x)\text{ d}x}&\ge b\int\limits_{a}^{b}{f(x)\text{ d}x} - \int\limits_{a}^{b}{(x-a)f(x)\text{ d}x} \\
&= (b+a)\int\limits_{a}^{b}{f(x)\text{ d}x} - \int\limits_{a}^{b}{xf(x)\text{ d}x}
\end{align*}
and thus
$$ 2\int\limits_{a}^{b}{xf(x)\text{ d}x}\ge(b+a)\int\limits_{a}^{b}{f(x)\text{ d}x}.$$
A: Just another way of stating it:
The inequality can be written
$$
\int_a^b f(x)\Bigl(x-\frac{a+b}{2}\Bigr)\,dx\geq 0.\tag{*}
$$
Since
$$
\int_a^b c\Bigl(x-\frac{a+b}{2}\Bigr)\,dx=0
$$
the claimed inequality $(*)$ is equivalent to
$$
\int_a^b \bigl(f(x)-c\bigr)\Bigl(x-\frac{a+b}{2}\Bigr)\,dx\geq 0,\tag{**}
$$
for any $c\in\mathbf R$. With $c=f((a+b)/2)$, the integrand in $(**)$ becomes nonnegative, and so the integral is clearly non-negative.
A: Let $y = a+b - x$, notice 
$$\int_a^b f(x) = \int_a^b f(y) dx\quad\text{ and }\quad\int_a^b x f(x) dx = \int_a^b yf(y) dx$$
We have
$$\begin{align}\int_a^b x f(x) dx 
&= \frac12 \int_a^b (x f(x) + y f(y) ) dx\\
&= \frac14 \int_a^b \bigg[
(\underbrace{x + y}_{= a+b})(f(x)+f(y)) + 
\underbrace{(x-y)(f(x)-f(y))}_{\ge 0  \text{ because } f \text{ is increasing }}
\bigg] dx\\
&\ge \frac{a+b}{4}\int_a^b (f(x)+f(y)) dx\\
&=   \frac{a+b}{2}\int_a^b f(x) dx
\end{align}
$$
A: Consider
$$
\begin{align}
0
&\color{#C00}{\le}\int_a^b\overbrace{\left(x-\frac{b+a}2\right)}^{\color{#090}{\substack{\text{integral vanishes}\\\text{over $[a,b]$}}}}\overbrace{\left(f(x)-f\left(\frac{b+a}2\right)\right)}^{\color{#C00}{\substack{\text{same sign as $x-\frac{b+a}2$}\\\text{since $f(x)$ is non-decreasing}}}}\,\mathrm{d}x\\
&\color{#090}{=}\int_a^b\left(x-\frac{b+a}2\right)f(x)\,\mathrm{d}x\\
&=\int_a^bxf(x)\,\mathrm{d}x-\frac{b+a}2\int_a^bf(x)\,\mathrm{d}x
\end{align}
$$
Thus,
$$
\int_a^bxf(x)\,\mathrm{d}x\ge\frac{b+a}2\int_a^bf(x)\,\mathrm{d}x
$$
