# Dan Sitaru's Inequality with Three Related Integrals and Derivatives

### Solution

We shall repeatedly use integration by parts.

\displaystyle \begin{align} \int_0^a f(x)dx\;&=\int_0^a x'f(x)dx=xf(x)\Bigr|_0^a-\int_0^a xf'(x) dx\\ &=-\int_0^a xf'(x) dx \end{align}

\displaystyle \begin{align} \Bigr(\int_0^a f(x) dx\Bigr)^2\;&=\Bigr(\int_0^axf'(x)dx\Bigr)^2\leq \int_0^a x^2 dx\cdot \int_0^a \Bigr(f'(x)\Bigr)^2dx\\ &=\frac{x^3}{3}\Bigr|_0^a\cdot \int_0^a \Bigr(f'(x)\Bigr)^2dx=\frac{a^3}{3}\int_0^a \Bigr(f'(x)\Bigr)^2 dx\\ &=3\Bigr(\int_0^a f(x) dx\Bigr)^2\leq a^3 \int_0^a \Bigr(f'(x)\Bigr)^2 dx \end{align}

\displaystyle \begin{align} \int_0^a f(x) dx\;&=-\int_0^a xf'(x) dx=-\int_0^a \Bigr(\frac{x^2}{2}\Bigr)'f'(x) dx\\ &=-\Bigr(\frac{x^2}{2}f'(x)\Bigr|_0^a-\int_0^a\frac{x^2}{2}f''(x)dx\Bigr)=\frac{1}{2}\int_0^ax^2f''(x) dx\\ &\leq \Bigr(2\int_0^a f(x)dx\Bigr)^2 =\Bigr(\int_0^a x^2 f''(x) dx\Bigr)^2\\ &\leq \int_0^a x^4dx\cdot \Biggl(\int_0^a \Bigr(f''(x)\Bigr)^2 dx\Biggl)\\ &=\frac{x^5}{5}\Bigr|_0^a \Biggl(\int_0^a \Bigr(f''(x)\Bigr)^2 dx\Biggl)\\ &=\frac{a^5}{5}\Biggl(\int_0^a\Bigr(f''(x)\Bigr)^2 dx\Biggl) \end{align}

\displaystyle \begin{align} 4\Biggl(\int_0^a f(x) dx\Biggl)^2&\leq \frac{a^5}{5}\Biggl(\int_0^a \Bigr(f''(x)\Bigr)^2 dx\Biggl)\\ 3\Biggl(\int_0^a f(x) dx\Biggl)^2&\leq a^3 \Biggl(\int_0^a \Bigr(f'(x)\Bigr)^2 dx\Biggl)\\ 12\Biggl(\int_0^a f(x) dx\Biggl)^4&\leq \frac{a^8}{5}\Biggl(\int_0^a\Bigr(f'(x)\Bigr)^2dx\Biggl)\Biggl(\int_0^a \Bigr(f''(x)\Bigr)^2 dx\Biggl)\\ \left(\int_0^af(x)dx\right)^4&\leq \frac{a^8}{60}\left(\int_0^a\left(f'(x)\right)^2 dx\right)\left(\int_0^a \left(f''(x)\right)^2 dx\right) \end{align}

### Acknowledgment

This is a Dan Sitaru's problem from the Romanian Mathematical Magazine. Dan has kindly sent me the problem and his solution on a LaTeX file. I very much apppreciate this kind of thoughtfulness. Amit Itagi has independently come up with the same solution.