16
$\begingroup$

Prove that

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

$\endgroup$
23
$\begingroup$

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

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

$\endgroup$
  • 3
    $\begingroup$ A very nice way! (+1) $\endgroup$ – user 1357113 Nov 6 '12 at 21:08
5
$\begingroup$

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

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

$\endgroup$
3
$\begingroup$

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 $$ 1-\frac{1}{6}x^2-\frac{1}{x+1}=\frac{1}{6}\cdot\frac{x(6-x^2-x)}{x+1}\geq 0. $$

$\endgroup$
  • $\begingroup$ (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
2
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.