An Integral Inequality from the RMM

The condition $f'(x)=f'(1-x)\;$ implies $f(x)+f(1-x)=C,\;$ a constant, for $x\in [0,1].$ This is a kind of situation that has been considered elsewhere on three different occasions.

If $\displaystyle I=\int_{0}^{1}f(x)dx,\;$ then

$\displaystyle 2I=\int_{0}^{1}(f(x)+f(1-x))dx=C=f(x)+f(1-x),\;$

for any $x\in [0,1].\;$ In particular, with $x=0,\;$ $2I=f(0)+f(1).\;$ It then follows by the AM-GM inequality that

$\displaystyle I=\frac{f(0)+f(1)}{2}\ge\sqrt{f(0)f(1)}.$

With $\displaystyle I=\int_{0}^{1}f(x)dx,\;$ and integrating by parts,

$\displaystyle\begin{align} I&= \left[xf(x)\right]^1_0-\int_0^1xf'(x)dx\\ &=f(1)-\int_0^1xf'(x)dx\\ &=f(1)+\int_0^1xf'(1-x)d(1-x)\\ &=f(1)+\int_0^1f'(1-x)d(1-x)-\int_0^1(1-x)f'(1-x)d(1-x)\\ &=f(1)+\int_1^0f'(x)dx-\int_1^0xf'(x)dx\\ &=f(1)+f(0)+\int_{1}^{0}\left[\frac{d}{dx}(x)\int_{1}^{0}f'(x)dx\right]dx\\ &=f(1)+f(0)-\int_0^1f(x)dx\\ &=f(1)+f(0)-I, \end{align}$

implying

$\displaystyle I=\frac{f(0)+f(1)}{2}\ge\sqrt{f(0)f(1)}.$

The problem from the Romanian Mathematical Magazine has been posted at the CutTheKnotMath facebook page by Dan Sitaru, along with the solution (Proof 1) by Safal Das Biswas (India). To avoid repetition I took the liberty of shortening the proof, using links to the older pages at this site. Proof 2 is by Diego Alvariz (India).