# An Inequality with Sin, Cos, Tan, Cot, and Some

### Solution

First for $x\in \Bigr(0,\frac{\pi}{2}\Bigr)$ we prove that:

$\sin x+\tan x\gt 2x.$

Let be $f:\Bigr(0,\frac{\pi}{2}\Bigr)\to \mathbb{R}; f(x)=\sin x+\tan x-2x.\,$ We have $f'(x)=\cos x+\tan^2x-1\,$ and $\displaystyle f''(c)=\frac{\sin x}{\cos^3 x}(2-\cos^2 x)\gt 0.\,$ It follows that

$\displaystyle f'(x)\gt\lim_{x\to 0^{+}} f'(x)=0$

such that $f'(x)\gt 0,\,$ $x\in\displaystyle\left(0,\frac{\pi}{2}\right),\,$ implying

$\displaystyle f(x)\gt \lim_{x\to 0^{+}}f(x)=0,$

so that $f(x)\gt 0,\,$ $x\in\displaystyle\left(0,\frac{\pi}{2}\right).\,$ Hence,

(1)

$\sin x+\tan x\gt 2x.$

Replace in (1) $x\,$ with $\displaystyle \frac{\pi}{2}-x$:

(2)

$\cos x+\cot x\gt 2\displaystyle\Bigr(\frac{\pi}{2}-x\Bigr).$

(3)

$\sin x+\tan x+\cos x+\cot x>\pi.$

For $x=A; x=B; x=C$ in (3) and adding up:

(4)

$\displaystyle\sum_{cycl}(\sin A+\cos A+\tan A+\cot A)\gt 3\pi.$

By the AM-GM inequality, $a+b+c\geq 3\sqrt[3]{abc}\,$ such that $\displaystyle 8\Bigr(\frac{a+b+c}{2}\Bigr)^3\geq 27abc,\,$ i.e., $8s^3\geq 27abc.\,$ Further $\displaystyle 8s^3\geq 27\cdot 4RS=27\cdot 4Rrs,\,$ or, $27\cdot Rr\leq 2s^2\,$ and, finally,

(5)

$\displaystyle 9Rr\leq \frac{2s^2}{3}.$

Now for a few facts, concerning the orthic triangle of $\Delta ABC:\,$ the side length of the orthic triangle are:

$a'=a\cos A, b'=b\cos B, c'=c\cos C$

The inradius: $r'=2R\cos A\cos B\cos C,$
The circumradius: $\displaystyle R'=\frac{R}{2},$
The semiperimeter: $s'=\displaystyle\frac{S}{R}$.

Now, we apply (1) to the orthic triangle:

\displaystyle\begin{align} &9R'r'\leq \frac{2s^2}{3},\\ &9\cdot \frac{R}{2}\cdot 2R \cos A\cos B\cos C\leq \frac{2}{3}\cdot \frac{S^2}{R^2},\\ &\cos A\cos B\cos C\leq \frac{2S^2}{27R^4},\\ &\frac{1}{\displaystyle\prod_{cycl}\cos A}\geq \frac{27R^4}{2S^2},\\ &\frac{2S^2}{\displaystyle\prod_{cycl}\cos A}\geq 27R^4, \end{align}

implying $2S^2\geq 27R^4\displaystyle\prod_{cycl}\cos A.\,$ Multiplying (4), (5):

$\displaystyle 2S^2\sum_{cycl}(\sin A+\cos A+\tan A+\cot A)\gt 81\pi R^4\prod_{cycl}\cos A.$

### Acknowledgment

I am grateful to Dan Sitaru for posting this problem from his book "Math Accent" at the CutTheKnotMath facebook page and later supplying a LaTeX file with its solution.