How to prove that $ \int_0^1 \frac{e^x}{x+1}dx\le \frac e2$ using the inner product.

Question

Let

$$\langle f,g \rangle=\displaystyle\int_0^1f(t)g(t)dt$$ be an inner product over $C[0,1]$. How to prove that

$$\displaystyle\int_0^1 \dfrac{e^x}{x+1}dx\le \dfrac e2$$

My miserable work: by Cauchy-Schwarz inequality $|\langle f,g \rangle|\le \|f\|\cdot\|g\|$, and if we were to take $f=e^x$, $g=1/(x+1)$ , observe that $f,g\in C[0,1]$

$$\displaystyle\int_0^1 \dfrac{e^x}{x+1}dx\le \sqrt{\displaystyle\int_0^1 e^{2x}dx} \sqrt{\displaystyle\int_0^1 \dfrac{1}{(x+1)^2}dx}=\sqrt{\frac{1}{2e^2}-\frac{1}{2}}\sqrt{\frac{1}{2}}\le \frac{e}{2}$$

Answer 1

Due to convexity of the integrand on $[0,1]$,

\begin{align} \int_0^1 \dfrac{\exp(x)}{x+1}dx &< \int_0^1 \Big((\tfrac{\mathrm{e}}2-1)\,x+1 \Big) \, dx =\frac12+\frac{\mathrm e}4 <\frac{\mathrm e}2 . \end{align}

And if you like, you can write

\begin{align} \int_0^1 \Big((\tfrac{\mathrm{e}}2-1)\,x+1 \Big) \, dx &\le \sqrt{\int_0^1 1\,dx} \cdot \sqrt{\int_0^1 \Big((\tfrac{\mathrm{e}}2-1)\,x+1 \Big)^2 \, dx} \\ &=\frac16\,\sqrt{3\,\mathrm{e}^2+6\,\mathrm{e}+12} <\frac{\mathrm e}2 . \end{align}