# Leo Giugiuc's Exercise

### Solution 1

If $x=0,\,$ there's nothing to prove as both sides vanish. So assume $\displaystyle c\in\left(0,\frac{\pi}{2}\right],\,$ the inequality is equivalent to

(*)

$\displaystyle 2\left(\frac{\sin x}{x}\right)^2-\left(\frac{\sin x}{x}\right)-\cos x\ge 0.$

By Taylor's expansions:

\displaystyle \begin{align} 2\left(\frac{\sin x}{x}\right)^2&\ge 2\left(1-\frac{x^2}{3}+\frac{2x^4}{45}-\frac{x^6}{315}\right)^2\\ -\frac{\sin x}{x}&\ge -1+\frac{x^2}{6}-\frac{x^4}{120}\\ -\cos x&\ge -1+\frac{x^2}{2}-\frac{x^4}{24}. \end{align}

$\displaystyle LHS(*)\ge \frac{4x^4}{45}-\frac{x^4}{20}-\frac{2x^6}{315}=x^4\left(\frac{7}{180}-\frac{2x^2}{315}\right)\ge 0.$

### Solution 2

Let $f(x)=2\sin^2x-x\sin x-x^2\cos x.\,$ We need to show that $f\ge 0.\,$ Dividing by $\sin x,\,$ this reduces to

$2\sin x-x-x^2\cot x\ge 0.$

We'll use the inequality by Lv Hui-Lin et al (Sharp inequalities for tangent functions with applications, J. of Inequalities and Applications, 2017.1, 94): for $0\lt t\lt \displaystyle \frac{\pi}{2},$

$\displaystyle 1-\frac{4t^2}{\pi^2}\lt \frac{t}{\tan t}\lt 1-\frac{t^2}{3}.$

Accordingly, $\displaystyle -\cot x\gt -\frac{\displaystyle 1-\frac{x^2}{3}}{\displaystyle x}.\,$ So we need to prove

$\displaystyle f_2(x)=-x+x\left(-1+\frac{x^2}{3}\right)+2\sin x\ge 0.$

Observe that the inequality holds for $\displaystyle x=\frac{\pi}{2}\,$ and as $x\to 0.\,$ We can prove that it holds in-between since $f_2\,$ is monotone positive in the interval, for $f'_2(x)=x^2+2\cos x-2\,$ which is positive for all values $\displaystyle x\in\left(0,\frac{\pi}{2}\right].$

### Extra

Prove that, for $\displaystyle x\in\left[0,\frac{\pi}{2}\right],$

$\displaystyle x^2\le \sin x\cdot\tan x.$

From $\displaystyle x\sin x+x^2\cos x\le 2\sin^2x,$ $x^2\le 2\sin x\cdot\tan x-x\tan x,\,$ but, since $\sin x\lt x,\,$ we have

\displaystyle \begin{align}x^2&\le 2\sin x\cdot\tan x-x\tan x\\ &= \sin x\cdot\tan x+ \tan x(\sin x-x)\\ &\le \sin x\cdot\tan x. \end{align}

### Acknowledgment

Leo Giugiuc has kindly reposted the above problem from the mathematical inequalities facebook group to the CutTheKnotMath facebook page. The problem is by Leo Giugiuc, Solution 1 is due to Sladjan Stankovik's; Solution 2 is by N. N. Taleb.

Mike Lawler made an interesting observation, hightlighted above as Extra.