Integral inequality $e^{\int _0^1f\left(x\right)dx-1\:}\le \frac{1}{4}\int _0^1\left(x+1\right)e^{f\left(x\right)}dx$

Question

Let $f:\left[0,1\right]\rightarrow\mathbb{R}$ be a continuous function. Show that: $$e^{\int _0^1f\left(x\right)dx-1\:}\le \frac{1}{4}\int _0^1\left(x+1\right)e^{f\left(x\right)}dx$$ How should this inequality be approached? It probably has a straightforward solution but I could use a hint.

Answer 1

In explicit terms: Jensen's inequality gives $$ e^{\int_{0}^{1}g(x)\,dx}\leq \int_{0}^{1}e^{g(x)}\,dx $$ and if we pick $g(x)=\log(x+1)+f(x)$ this gives $$ e^{\log\frac{4}{e}+\int_{0}^{1}f(x)\,dx} \leq \int_{0}^{1}(x+1) e^{f(x)}\,dx $$ or $$ \frac{4}{e}\cdot e^{\int_{0}^{1}f(x)\,dx} \leq \int_{0}^{1}(x+1) e^{f(x)}\,dx $$ or $$ e^{\int_{0}^{1}f(x)\,dx-1} \leq \frac{1}{4}\int_{0}^{1}(x+1) e^{f(x)}\,dx $$ as wanted.