Let $f:[0,1]\to \mathbb{R}$ be non decreasing function. Show that following inequality is true: $$3\int^{1}_{0}f(x)\sqrt{x}dx\ge2\int^{1}_{0}f(x)dx$$
What I only came up with is that mean value for $\sqrt{x}$ at $[0,1]$ is $\frac{2}{3}$ but it doesn't seem as if I could replace $\sqrt{x}$ with that under integral sign.
What to do next? Any help will be appreciated.
The inequality is equivalent to $$\int_0^1 f(x) \left(\sqrt{x}-\frac{2}{3}\right) dx \geq 0.$$ Denote $g(x) = \sqrt{x}-\frac{2}{3}$. We know $g(x)$ is increasing over $[0,1]$ and $\int_0^1 g(x) = 0$. Obviously, $g(0)<0$ and $g(1)>0$, so there is a unique point $x_0 \in [0,1]$ s.t. $g(x_0) = 0$. Notice that $g(x)$ is non-positive over $[0,x_0]$ and non-negative over $[x_0,1]$. Along with the fact that $f(x)$ is non-decreasing, we have \begin{aligned} \int_0^1 f(x)g(x)dx &= \int_0^{x_0} f(x)g(x)dx + \int_{x_0}^1 f(x)g(x)dx \\ &\geq \int_0^{x_0} f(x_0)g(x) dx + \int_{x_0}^1 f(x_0)g(x)dx \\ &= f(x_0) \int_0^1 g(x) dx \\ &=0. \end{aligned}