A problem about center of mass

Question

Suppose $f(x)$ is positive, increasing and Riemann-integrable on the interval $[a,b]$. Let$$\bar{x}=\frac{\int_a^b xf(x)\,\text{d}x}{\int_a^b f(x)\,\text{d}x}.$$Prove that$$\int_a^{\bar{x}} f(x)\,\text{d}x \le \int_{\bar{x}}^b f(x)\,\text{d}x.$$I tried the mean value theorem, but failed to reach the conclusion. Can anyone give a hint?

Answer 1

Let us reduce the question to a known result. Replacing $f$ with $(\int_a^bf)^{-1}f$ if necessary,we may, (and will), suppose without loss of generality that $\int_a^bf=1$.

For every convex function $\varphi$ we have by Jensen's inequality $$\varphi\left(\int_a^b xf(x) dx\right)\leq \int_a^b\varphi(x)f(x)dx$$ or, using the notation of the problem: $$\varphi\left(\bar{x}\right)\leq \int_a^b\varphi(x)f(x)dx$$ Now, since $f$ is increasing then $\varphi:[a,b]\to\Bbb{R}:\varphi(x)=\int_a^xf(t)dt$ is convex, and consequently, we can apply the previous inequality with this $\varphi$: $$\int_a^{\bar{x}}f(t)dt\leq \int_a^b\left(\int_a^xf(t)dt\right)f(x)dx =\left.\frac{1}{2}\left(\int_a^xf(t)dt\right)^2\right\vert_{x=a}^{x=b}=\frac{1}{2}$$ and the desired inequality follows since $$\int_a^{\bar{x}}f(t)dt+\int_{\bar{x}}^bf(t)dt=1$$ the desired inequality follows.$\qquad\square$