Prove that

$$ \sin x \ge \frac{x}{x+1}, \space \space\forall x \in \left[0, \frac{\pi}{2}\right]$$


Take $x \in [0, \pi/2]$. Consider the right triangle with sides $1, x$ and $\sqrt{1 + x^2}$. The angle opposite the side with length $x$ is smaller than $x$. It follows that

$$ \sin(x) \geq \frac{x}{\sqrt{x^2 + 1}} \geq \frac{x}{x + 1}. $$

  • 3
    $\begingroup$ A very nice way! (+1) $\endgroup$ – user 1357113 Nov 6 '12 at 21:08

On the given interval:

$$f(x):=(x+1)\sin x-x\Longrightarrow f'(x)=\sin x+(x+1)\cos x-1\geq 0$$

since $\,\sin x+(x+1)\cos x\geq 1$ on $\,[0,\pi/2]\,$.

Thus, $\,f\,$ is monotone non-descending on $\,\left[0,\dfrac{\pi}{2}\right]\,$ and thus

$$ f(x)=(x+1)\sin x-x\geq 0=f(0)$$


I proved earlier (see my answer for $\lim_{x \to 0}{\frac{x-\sin{x}}{x^3}}=\frac{1}{6}$) that $$ \sin\,x\geq x-\frac{1}{6}x^3,\quad x\in[0,\pi/2]. $$ From this $$ \frac{\sin\,x}{x}\geq 1-\frac{1}{6}x^2. $$ Now for $x\in[0,\pi/2]$ we have $$ 1-\frac{1}{6}x^2-\frac{1}{x+1}=\frac{1}{6}\cdot\frac{x(6-x^2-x)}{x+1}\geq 0. $$

  • $\begingroup$ (It might be worth explaining that last inequality in a bit more detail - it took me a moment to realize that $6-x^2-x$ has to be manifestly positive because of the domain of $x$.) $\endgroup$ – Steven Stadnicki Nov 6 '12 at 20:38
  • $\begingroup$ @StevenStadnicki Thanks, I added. $\endgroup$ – vesszabo Nov 6 '12 at 20:41

Inequality is easily verified for $x=0,1$

$\csc x - \cot x = \frac{1}{\csc x + \cot x}<1 $ for $x \in \left(0, \frac{\pi}{2} \right)$

Hence, $\csc x < 1 +\cot x < 1+ \frac{1}{x} $ since its well known that $\tan x>x$

and this leads to the desired inequality


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