8
$\begingroup$

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 $$ \cos(x_1)+\dotsb+\cos(x_n) \le \frac{n\sqrt 2}2? $$

$\endgroup$
  • $\begingroup$ 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
6
$\begingroup$

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 $x_n\rightarrow\frac{\pi}{2}^-$

$\endgroup$
  • $\begingroup$ 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
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.