It is easily seen that if $x,y\in[0,\pi/2)$ satisfy $\tan(x)\tan(y)=1$, then $$\cos(x)+\cos(y)\le\sqrt 2$$

A much more delicate fact is that if $\tan(x)\tan(y)\tan(z)=1$ (while $0\le x,y,z<\pi/2$), then $$\cos(x)+\cos(y)+\cos(z)\le \frac{3\sqrt 2}2$$ I can prove this, but the proof is a little complicated; can anybody suggest a nice, simple proof?

As a generalization, suppose that $n\ge 4$, $x_1,\dotsc,x_n\in[0,\pi/2)$, and $\tan(x_1)\dotsb\tan(x_n)=1$. Does it follow that $$ \cos(x_1)+\dotsb+\cos(x_n) \le \frac{n\sqrt 2}2? $$

  • $\begingroup$ Could you provide your proofs? $\endgroup$ – Dr. Mathva Feb 13 at 12:15
  • $\begingroup$ You can try first to rewrite the cosine using cos^2=1/(1+tg^2). Than a=tg(x),b=tg(y), c=tg(x), the costraint becomes abc=1 with a,b,c>0. Than go on with e^a'=a, e^b'=b and e^c'=c, so that the constrain becomes additive, a'+b'+c'=0. Using convexity finally of f(x)=\sqrt(1/(1+e^2x) should do the job. It is just an idea, but I do not have time to see if it really works :) $\endgroup$ – Thomas Feb 13 at 12:17
  • $\begingroup$ @Thomas I checked your idea. It does not work. $\endgroup$ – Michael Rozenberg Feb 13 at 12:20
  • $\begingroup$ @Dr.Mathva: $n=2$ is almost immediate, for $n=3$ see Michael Rozenberg's solution below. $\endgroup$ – W-t-P Feb 13 at 12:25
  • $\begingroup$ @Michael Rozenberg . Yes that function is not convex. It would have been too easy :) Thx a lot for checking! $\endgroup$ – Thomas Feb 13 at 12:31

For three variables we ca use C-S:

Let $\tan{x}=\sqrt{\frac{b}{a}},$ $\tan{y}=\sqrt{\frac{c}{b}},$ where $a$, $b$ and $c$ are positives.

Thus, $\tan{z}=\sqrt{\frac{a}{c}}$ and by C-S we obtain: $$\sum_{cyc}\cos{x}=\sum_{cyc}\frac{1}{\sqrt{1+\tan^2x}}=\sum_{cyc}\sqrt{\frac{a}{a+b}}\leq$$ $$\leq\sqrt{\sum_{cyc}\frac{a}{(a+b)(a+c)}\sum_{cyc}(a+c)}\leq\frac{3}{\sqrt2},$$ where the last inequality it's just $$\sum_{cyc}c(a-b)^2\geq0.$$

The generalization is wrong for all $n\geq4$.

Try $x_1=x_2=...=x_{n-1}\rightarrow0^+$ and $x_n\rightarrow\frac{\pi}{2}^-$

  • $\begingroup$ This is exactly the solution for $n=3$ I had in mind (actually, learned it from you). $\endgroup$ – W-t-P Feb 13 at 12:24
  • $\begingroup$ @W-t-P Your generalization is very interesting. I'll think about this later because I am very very busy this week. Sorry. $\endgroup$ – Michael Rozenberg Feb 13 at 12:28

After the substitutions described in the comments ($cos(x_i) \rightarrow \sqrt{\frac{1}{1+tg^2(x_i)})}$ and $e^{a_i}=tg(x_i)$, one sees that needs to maximize:

$f(x)=\sum_{i=1}^{n} \sqrt{\frac{1}{1+e^{2x_i}})}$

subject to:

$\sum_i x_i=0$

over the domain $x \in R^n$.

This sounds like a job for Lagrange mulipliers but I was not able to solve the system. For the moment I observe that the conjecture is equivalent to having maximum for $x_i=0$ and I want to show that this is not the case for large $n$. Indeed $f(0,...0)=n\sqrt{2}/2$, but $f(-m,-m,...,-m,(n-1)m)\rightarrow (n-1)$ for large m. The second term dominates for large $n$ and therefore the thesis cannot be true.

  • $\begingroup$ Too bad I just saw that the counter example had already been proposed editing the answer by Michel Rosenberg... Well since I did the effort to write my answer I will leave it... $\endgroup$ – Thomas Feb 14 at 8:17
  • $\begingroup$ Anyway this "symmetry breaking" as a function of n (dimensionality) is very nice... Looks a bit "physical" :) $\endgroup$ – Thomas Feb 14 at 8:29

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