Showing Lebesgue Integral inequalities

Question

Let $f,g: [0,1] \longrightarrow (0,\infty)$ be measurable and $\beta >0$. Assume that $$\int_{0}^{1}g(x)dx = 1.$$ Show that $$1\leq \Bigg(\int_{0}^{1}f(x)^{-\beta}g(x)dx\Bigg)\Bigg( \int_{0}^{1}f(x)g(x)dx\Bigg)^{{\beta}}.$$ My attempt: Consider the measure $\mu$ defined by $$\mu(E)=\int_Eg(x)dx,$$ whenever $E$ is lebesgue measurable subset of $[0,1].$ From here im stuck, i think Jensen's inequality is useful here but i dont know how should apply it.

Any help and hints would be appreciated!!

Answer 1

$h(t)=t^{-\beta}$ is a convex function on $(0,\infty)$ because its second derivative is positive. Hence $h(\int f d\mu) \leq \int h(f)d\mu$ This gives the desired inequality.