# If $\tan(x_1) \cdots\tan(x_n)=1$ for acute $x_i$, then does it follow that $\cos(x_1)+\cdots+\cos(x_n) \leq n\sqrt{2}/2$?

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

• Could you provide your proofs? – Dr. Mathva Feb 13 at 12:15
• 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 :) – Thomas Feb 13 at 12:17
• @Thomas I checked your idea. It does not work. – Michael Rozenberg Feb 13 at 12:20
• @Dr.Mathva: $n=2$ is almost immediate, for $n=3$ see Michael Rozenberg's solution below. – W-t-P Feb 13 at 12:25
• @Michael Rozenberg . Yes that function is not convex. It would have been too easy :) Thx a lot for checking! – Thomas Feb 13 at 12:31

## 2 Answers

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

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

• 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... – Thomas Feb 14 at 8:17
• Anyway this "symmetry breaking" as a function of n (dimensionality) is very nice... Looks a bit "physical" :) – Thomas Feb 14 at 8:29