# An Inequality in Triangle, with Integrals

### Solution

By Kober's inequality, $\displaystyle \cos x\ge 1-\frac{2x}{\pi}.\,$ It follows that $\displaystyle \cos (\sin x)\ge 1-\frac{2\sin x}{\pi}\,$ such that, say,

\displaystyle\begin{align} \int_0^a\cos (\sin x)dx&\ge\int_0^a\left(1-\frac{2\sin x}{\pi}\right)dx\\ &=a+\frac{2}{\pi}\cos x\bigg|_0^a=a+\frac{2}{\pi}(\cos a-1)\\ &=a+\frac{2}{\pi}\left(1-2\sin^2\frac{a}{2}-1\right)\\ &=a-\frac{4}{\pi}\sin^2\frac{a}{2}. \end{align}

Thus, altogether,

\displaystyle\begin{align}4\sum_{cycl}\sin^2\frac{a}{2}&+\pi\sum_{cycl}\int_{0}^{a}\cos (\sin x)dx\\ &\ge 4\sum_{cycl}\sin^2\frac{a}{2}+\sum_{cycl}\left(\pi a-4\sin^2\frac{a}{2}\right)\\ &=\pi\sum_{cycl}a=\pi^2. \end{align}

### Refinement

For $\displaystyle a,b,c\in \left[0,\frac{\pi}{2}\right],\,$ $a+b+c=\pi,\,$ prove that

$\displaystyle \frac{\pi}{3}\sum_{cycl}\sin^2\frac{a}{2}+\sum_{cycl}\int_{0}^{a}\cos (\sin x)dx\ge \pi.$

It's known that $\displaystyle \sum_{cycl}\sin^2\frac{a}{2}\ge\frac{3}{4}\,$ and $\displaystyle \cos t\ge 1-\frac{t^2}{2},\,$ $t\gt 0.\,$ From the latter, $\displaystyle \cos (\sin x)\ge 1-\frac{\sin^2x}{2},\,$ for $\displaystyle x\in \left[0,\frac{\pi}{2}\right].$

Suffice it to show that $\displaystyle \sum_{cycl}\int_{0}^{a}\left(1-\frac{\sin^2x}{2}\right)dx\ge\frac{3\pi}{4}.\,$ But

\displaystyle \begin{align} \sum_{cycl}\int_{0}^{a}\left(1-\frac{\sin^2x}{2}\right)dx &= \pi-\sum_{cycl}\int_{0}^{a}\left(\frac{\sin^2x}{2}\right)dx\\ &=\pi+\sum_{cycl}\left(-\frac{a}{4}+\frac{\sin 2a}{8}\right)\\ &\ge \pi-\frac{\pi}{4}=\frac{3\pi}{4}. \end{align}

### Acknowledgment

This is Dan Sitaru's problem from the Romanian Mathematical Magazine. Solution is by Dan Sitaru. Leo Giugiuc came up with an essential refinement above.