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I was just experimenting with WolframAlpha today and noticed that the solution of the inequality $\sec^{-1} x\tan^{-1} x$ is $x\leq1$. I tried to prove the inequality for $x\geq1$ to be false.

My attempt:

Let $h(x)=\sec^{-1} x-\tan^{-1} x$. We have to prove that $h(x)<0 \forall x\geq1$. Note that $h(1)=-\pi/4$.

If $h(x)$ is a decreasing function, then QED already.
If $h(x)$ is an increasing function, then we note that $\lim _{x\to\infty}h(x)=\pi/2-\pi/2=0$. Hence, QED.

Both cases above are monotonic because the functions $\sec^{-1} x$ and $\tan^{-1} x$ are themselves monotonic.


Is my proof correct?

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  • $\begingroup$ Of course $h(x)$ can be either increasing or decreasing only, but I noticed that both cases satisfy the inequality, so I was too lazy to delve into the double derivative :P $\endgroup$ Dec 30 '17 at 2:18
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You want to prove that $$\sec^{-1} x<\tan^{-1} x\;\;\;\forall x\geq1$$

To prove this, let $y=\sec^{-1}x$, $z=\tan^{-1}x$. $$x=\frac1{\cos y}=\frac{\sin z}{\cos z}\\\implies\cos z=\sin z\cos y$$ Since we are looking at $x\ge1$, this means that $\sin z\ge\cos z$, i.e. $z\in[\pi/4,\pi/2)$, and $y\in[0,\pi/2)$.

Now fix $y$. Then $$\cos z=\cos y \sin z\\\implies |\cos z|=|\cos y|\cdot|\sin z|<|\cos y|\\\implies\cos z<\cos y$$since both are positive. Since $\cos$ is decreasing on this interval, this implies that $y<z$, i.e. $$\sec^{-1}x<\tan^{-1}x$$

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Both cases above are monotonic because the functions $\sec^{-1} x$ and $\tan^{-1} x$ are themselves monotonic.

The sum of two monotonic functions is not necessarily monotonic. Try out $f(x)=\sin x-x, \ g(x)=\cos x+x$ and you see that $f(x)+g(x)$ is not monotonic. Same applies to the difference of two monotonic functions.

Thus, you cannot establish the monotonicity of $h(x)=\sec^{-1} x-\tan^{-1} x$ that way in your proof.

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  • $\begingroup$ Thanks for your answer! But, it seems like both of your examples are indeed monotonic. The points where they have first derivative zero happens to be an inflexion point. $\endgroup$ Dec 30 '17 at 3:49
  • $\begingroup$ @GaurangTandon If you want avoid this, you can construct functions like y=sin(x)+(x-k)^3, k>0... hmm and there shouldn't be such inflection points. $\endgroup$
    – Macrophage
    Dec 30 '17 at 4:01
  • $\begingroup$ Yes, that's better. $\endgroup$ Dec 30 '17 at 4:39

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