Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Prove that

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

share|improve this question
add comment

3 Answers 3

up vote 19 down vote accepted

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

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

share|improve this answer
A very nice way! (+1) –  Chris's sis Nov 6 '12 at 21:08
add comment

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

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

share|improve this answer
add comment

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

share|improve this answer
(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 $x$.) –  Steven Stadnicki Nov 6 '12 at 20:38
@StevenStadnicki Thanks, I added. –  vesszabo Nov 6 '12 at 20:41
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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