Proving $tan^{-1}frac{1}{x} leq frac{1}{x}$

Question

I'm not sure what I'm missing. I've done a limit comparison test that relies on $$tan^{-1}frac{1}{x} leq frac{1}{x}$$ which seems to be true for $x > 0$, but I'd like to prove it before letting my solution rely on it.
I've tried a few different things that led nowhere. Never faced a problem like this so I'm really not sure what's on the plate for strategies. (I'm in Calc 2.)
Thanks for any advice or nudges in the right direction!

user839911 · Answer

We put $f(x) =arct(frac{1}{x}) - frac{1}{x} $ f is derivable at $mathbb{R^{+*}} $

$f'(x) =frac{-frac{1}{x^2}}{1+frac{1}{x^2}}+frac{1}{x^2} =frac{1}{x^2}(1-frac{1}{1+x^2}) =frac{1}{1+x^2}>0 $

So :

$(*) f(x)$ is an incremental function at $mathbb{R^+} $

$(**) lim _{x to 0^+} f(x) =-infty$

$(***) lim _{x to +infty} f(x)=0$

After $(*) $ and $(**) $ and $(***) $ we can see

$f(x) = arct(frac{1}{x}) - frac{1}{x} <0$

Finally :

$forall x>0 $ $ arct(frac{1}{x}) < frac{1}{x}$

xpaul · Answer

You may like this method:
$$ arctan(frac1x)=int_0^{frac1x}frac1{1+t^2}dtleint_0^{frac1x}dt=frac1x. $$

TheSilverDoe · Answer

You are right, $arctan(x)<x$ for all $x>0$ (and you can apply this to $1/x$ to get your inequality).
Indeed $f:x mapsto x-arctan(x)$ is differentiable over $mathbb{R}_+^*$ and
$$f'(x)=1-frac{1}{1+x^2} >0$$
so $f$ is strictly increasing. And $f(0)=0$, so $f(x) > 0$ for all $x >0$. So $x > arctan(x)$.

Proving $tan^{-1}frac{1}{x} leq frac{1}{x}$

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