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I must be a terrible googling searcher but I cannot find a reference to the following inequality:

$$ \forall_{\phi\in(0;\frac \pi 4)}\ \ln(\cot(\phi)))\, <\, \cot(2\!\cdot\!\phi) $$

I have just obtained this, it seems to have nice potential, and now I would appreciate a reference. (As a minimum, if this is new to you, I hope that you enjoed it).

REMARK: I apologize for my trivial question; I'd be more than willing to remove it once I get a respective link or a paper.

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Plug $x=\cot(\phi)$ and turn your inequality into $$ x \ln x < \frac{x^2-1}{2},$$ where $x>1$. The RHS is the second order Taylor approximation of the LHS around $x=1$. Hence the inequality follows from concavity of the derivative of $x \ln x$, which is $1+\ln x$.

This argument has led me to the following elegant solution: Integrate the following well-known inequality between 1 and $x$:

$$ \ln t \le t-1,$$ and get $$ x \ln x - x + 1 \le \frac{(x-1)^2}{2} \implies $$ $$x\ln x \le \frac{x^2-1}{2}.$$ (Equality iff $x=1$).

[EDIT] Turns out there's a reference for the inequality $\log(1+x) < x-\frac{x^2}{2(1+x)}$, which is equivalent to your inequality.

Hermite–Hadamard inequality states the following: If $f:[a,b]\to \mathbb{R}$ is convex and continuous, then $$f\left( \frac{a+b}{2}\right) \le \frac{1}{b - a}\int_a^b f(x)\,dx \le \frac{f(a) + f(b)}{2}.$$ The proof is geometric - one compares the integral to a trapezoid and to a rectangle. Applying this to $f(x)=-\frac{1}{x}$ (as you have done below), one obtains the desired inequality.

In fact, this was the first application of this inequality. In the letter "Sur deux limites d'une intégrale d´e finie", published by Hermite in Mathesis 3 (1883, p. 82), he proves the above inequality and applies it to $f(x)=\frac{1}{x+1}, a=0$ and gets $$x −\frac{x^2}{2+x^2} < \log(1+x) < x-\frac{x^2}{2(1+x)}.$$

I was not able to locate the letter myself, but the relevant excerpt appears right in the beginning of the introduction to this monograph by Dragomir and Pearce.

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  • $\begingroup$ Hm, thank you. I already had a proof, of course, but my question was about a reference. I'll up-vote you anyway. I'll post my original proof too. These proofs are like walking from a vertex of a rectangle on the edges toward the opposite vertex, you go either East then North, or first you go North, and then East. $\endgroup$
    – Włodzimierz Holsztyński
    Commented Dec 25, 2015 at 7:28
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    $\begingroup$ Since this is a consequence of a famous inequality ($\ln x \le x-1$), I will be (pleasently) surprised to discover a more specific reference. Looking forward to read your proof! $\endgroup$
    – Ofir Gorodetsky
    Commented Dec 25, 2015 at 7:33
  • $\begingroup$ Well, all properties of the real logarithm follow from the homomorphic property and that one inequality (simple axioms). $\endgroup$
    – Włodzimierz Holsztyński
    Commented Dec 25, 2015 at 7:53
  • $\begingroup$ Let me be a little pedantic, sorry. The inequality, $\ \log(x) \le x-1\ $ is only logically strong since everything else follows. However, mathematically speaking, in the everyday mathematical sense, the straight trapezoid provides an immediately sharper inequality than the rectangular $\ \log(x) \le x-1\ $. $\endgroup$
    – Włodzimierz Holsztyński
    Commented Dec 25, 2015 at 8:05
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Looking at the integral of $\frac 1x$, i.e. at the respective curvilinear trapezoid, and at the straight polygonal trapezoid, one gets:

$$ \ln(b) - \ln(a)\ <\ \frac 12\cdot\left(\frac 1a+\frac 1b\right)\cdot(b-a) $$

whenever $\ 0<a<b;\ $ i.e.

$$ \ln\left(\frac ba\right)\ <\,\ \frac {b^2-a^2}{2\cdot a\cdot b} $$

Now let $\ 0<\phi<\frac \pi 4.\ $ Let $\ a:=\sin(\phi)\,$ and $\, b:=\cos(\phi),\ $ so that $0<a<b$. Then indeed we get the required inequality:

$$ \ln(\cot(\phi))\ <\ \cot(2\cdot \phi) $$

Have fun! This is the time of the year... ok, any time is great.

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    $\begingroup$ Nice proof! Now when I see the proof I think it might be possible to find a reference. For starters, this follows from the error analysis in the Trapezoidal rule: en.wikipedia.org/wiki/Trapezoidal_rule $\endgroup$
    – Ofir Gorodetsky
    Commented Dec 25, 2015 at 8:03
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    $\begingroup$ The Trapezoidal rule applied to convex\concave functions yields an inequality known as "Hermite–Hadamard inequality". In the first paper (by Hermite) which proves this inequality, he applies it to $f(x)=\frac{1}{x+1}$ and obtains your inequality. See my edit to my original answer. $\endgroup$
    – Ofir Gorodetsky
    Commented Dec 25, 2015 at 8:47
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    $\begingroup$ @OfirGorodetsky -- once again, many-many thanks. I consider writing elementary articles for amateurs and youngsters. I will certainly make a reference to He-Ha (and will mention that the info came to me from you). $\endgroup$
    – Włodzimierz Holsztyński
    Commented Dec 25, 2015 at 9:09
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    $\begingroup$ My pleasure. I enjoy learning about 19th century papers. I also want to correct myself - one side of the He-Ha inequality is by the "trapezoid rule", but the other side is by the "midpoint rule". $\endgroup$
    – Ofir Gorodetsky
    Commented Dec 25, 2015 at 9:19
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    $\begingroup$ @OfirGorodetsky -- I expected that the other side estimate would call for a different approach (naturally, due to the nature of the convexity). I was about to write that other side inequalityt but MO slowed me down (LaTeX etc takes a lot of energy, especially on MO). I'd call the other approach "tangential" or "derivative". It's 4:34am now. But possibly later I may add the other side inequality (no firm promises, just a vague expectation :-) $\endgroup$
    – Włodzimierz Holsztyński
    Commented Dec 25, 2015 at 9:34

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