inequality questions involving $\arctan$ and $\ln$

Question

I have two questions related to inequality:

1) Show that for all $x,y\in\mathbb{R}\thinspace\thinspace\thinspace |\arctan(x) - archtan(y)| \leq |x-y|\thinspace\thinspace$ I tried to use the triangle inequality and claim that $|\arctan(x) - arctan(y)| \leq |\arctan(x)| + |\arctan(y)|$ and try to show that $|\arctan(x)| + |\arctan(y)| \leq |x-y|$

2) Why for all $\; x>0$, $\;\ln(x) + \frac{1}{x} \geq0\,?$ How can I show this?

If archtan is a misprint of arctan, then use mean value theorem. — Angina Seng, Dec 29 '19 at 11:17
Please restrict your question to a single problem. – The first one is answered here: https://math.stackexchange.com/q/341741/42969 — Martin R, Dec 29 '19 at 11:17
@LordSharktheUnknown Yeah, just observed that, so deleted that comment. — 19aksh, Dec 29 '19 at 11:20
For the second inequality set $x=\dfrac{1}{X}$ : you have now to prove that $-\ln(X)+X>0$... Isn't it evident for any $X>0$ ? — Jean Marie, Dec 29 '19 at 11:23
Sorry for the typo, re-edited it. about the second need more detailed explanation, please. — HSHA, Dec 29 '19 at 11:27
I've never seen it called $\operatorname{archtan}$ before, only $\operatorname{artanh}$ or $\operatorname{arctanh}$. — J.G., Dec 29 '19 at 11:33

score 0 · Answer 1 · edited Dec 29 '19 at 12:12

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1) The inequality for $\arctan$ is immediate from MVT.

2): $\frac d {dx} (\ln x+\frac 1 x)=\frac 1x -\frac 1 {x^{2}} $ which is negative for $x <1$ and positive for $x>1$. Hence $\ln x+\frac 1 x$ is decreasing in $(0,1)$ and increasing in $(1,\infty)$ so its minimum value is attained at $x=1$. The minimum value is $1$ so we actually get $\ln x+\frac 1 x \geq 1$ for all $x >0$.

edited Dec 29 '19 at 12:12

answered Dec 29 '19 at 11:37

Kavi Rama Murthy

311,013

See as well the elementary solution I propose in a comment. – Jean Marie Dec 29 '19 at 11:39
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@JeanMarie OP is asking for a detailed proof. I suspect that OP would ask for details for $-\ln x +x >0$ also. – Kavi Rama Murthy Dec 29 '19 at 11:44

score 0 · Answer 2 · answered Dec 29 '19 at 12:24

(b) Considering only x>0, since y = 1/x has a line of symmetry y = x, the curve of y=1/x going up (near the y-axis) is the "same" as the curve of y = 1/x going right (near the x-axis). So question (b) is the same as: prove that e^-x < 1/x (1) for all x > 0.

But for all real a > 0 and b > 0, a < b iff 1/a > 1/b (See something like Rudin's PMA Chapter 1 for a proof of this).

So (1) is the same as: prove that e^x > x for all x > 0. For this, see here:

How to prove that $e^x=x$ has no real solution?

inequality questions involving $\arctan$ and $\ln$

2 Answers2