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I have to prove that $$\sin\left(\frac{1}{x}\right) < \frac{1}{x}$$ For all $x>0$

Usually what I do in these cases:

Let $f(x)=\sin\left(\frac{1}{x}\right) - \frac{1}{x}$

And now prove that $f(x)<0$ for every $x>0$.

$$f'(x)=\frac{\cos\left(\frac{1}{x} \right)-1}{x^2}$$ If I prove that $f'(x)$ <0, it would mean that the original function is decreasing which allows me the prove the statement.

I have found out that $f'(x)=0 \Leftrightarrow x=\frac{1}{2\pi}$.

However, also $f''(x)=0 \Leftrightarrow x=\frac{1}{2\pi}$.

I thought that if $x=\frac{1}{2\pi}$ is a minimum I can prove that statement based on that, but I can't manage to prove that.

Any Ideas?

Community
  • 1
Alan
  • 2,791

3 Answers3

5

If $y>1$, we automatically have $sin(y)<y$

Let $1\geq y>0$

Then by the mean value theorem,

$sin(y)= cos(c)y< y$ for some $c\in (0,y)$

Now, since the inequality holds for all $y>0$, we have if $x>0$, then

$sin(1/x)<1/x$

2

Substitute $y:=\frac{1}{x}$. Then your inequality is equivalent to $\sin(y)<y$ for all $y>0$. The classic proof for this uses the mean value theorem, see Ben Pineau's answer. Here is one using your approach.

Define $f(y):=\sin(y)-y$. Then $f'(y)=\cos(y)-1$. Since $\cos(y)\in[-1,1]$ for all $y\in\mathbb R$, we have $f'(y)\leq 0$, so $f$ is decreasing. Thus, $f(y)\leq f(0)=0$ and hence $\sin(y)\leq y$ for all $y\geq 0$. If $\sin(y_0)=y_0$ for some $y_0>0$, we had $f(y_0)=0$ and hence $f(y)=0$ for all $y\in(0,y_0)$. But then $f'(y)=\cos(y)-1=0$ for those $y$, which is impossible. Consequently, $\sin(y)<y$ for all $y>0$.

sranthrop
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Are you assuming that $x$ is positive? You really should, since it's not true when $x$ is negative. So you should be proving that $\sin y<y$ for all $y>0$.

This is a very well-known calculus result: a pretty standard proof uses the Mean Value Theorem,

Angina Seng
  • 158,341