Prove that $\int_a^bxf(x)dx\geq\frac{b+a}{2}\int_a^bf(x)dx$

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.

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.

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 $$ \int_a^bx\,f(x)\,\mathrm{d}x\ge\frac{b+a}2\int_a^b\,f(x)\,\mathrm{d}x $$