Prove $\tan x + \tan y + \tan z \le \frac{17}{6}$ for reals $\sum\limits_{\mathrm{cyc}} \sin x = 2, \sum\limits_{\mathrm{cyc}} \cos x = \frac{11}{5}$

Question

Problem. Let $x, y, z$ be real numbers with $\sin x + \sin y + \sin z = 2$ and $\cos x + \cos y + \cos z = \frac{11}{5}$. Prove that $$\tan x + \tan y + \tan z \le \frac{17}{6}.$$

I found this inequality when I dealt with this question: If $\sin x+\sin y+\sin z=2$, $\cos x+\cos y+\cos z=11/5$, $\tan x+\tan y+\tan z=17/6$, $x,y,z\in\mathbb{R},$ find $\sin(x+y+z)$ without a calculator

In detail, I let $\sin x = \frac{2a}{1 + a^2}, \cos x = \frac{1 - a^2}{1 + a^2}, 0 \le a < 1$ etc. and use Mathematica to find the maximum and minimum of $\frac{2a}{1-a^2} + \frac{2b}{1 - b^2} + \frac{2c}{1 - c^2}$ subject to $\frac{2a}{1 + a^2} + \frac{2b}{1 + b^2} + \frac{2c}{1 + c^2} = 2$ and $\frac{1 - a^2}{1 + a^2} + \frac{1 - b^2}{1 + b^2} + \frac{1 - c^2}{1 + c^2} = \frac{11}{5}$. Mathematica output: The maximum is simply $\frac{17}{6}$ when $a = 1/3, b = 1/3, c = 1/2$. The minimum is $984703/350796$ when $a = 9/32, b = 19/43, c = 19/43$.

I want to see a human verifiable proof.

Answer 1

I hope the following will help.

We need to prove that: $$\sum_{cyc}\tan{x}-\frac{17}{6}+\frac{125}{84}\left(\sum_{cyc}\sin{x}-2\right)+\frac{4625}{1008}\left(\sum_{cyc}\cos{x}-\frac{11}{5}\right)\leq0.$$ Now, by your idea let $\tan\frac{x}{2}=a,$ $\tan\frac{y}{2}=b$ and $\tan\frac{z}{2}=c$, where $\{a,b,c\}\subset(0,1).$

Thus, we need to prove that: $$\sum_{cyc}\frac{2a}{1-a^2}-\frac{17}{6}+\frac{125}{84}\left(\sum_{cyc}\frac{2a}{1+a^2}-2\right)+\frac{4625}{1008}\left(\sum_{cyc}\frac{1-a^2}{1+a^2}-\frac{11}{5}\right)\leq0,$$ which will true if we will prove that: $$\frac{2a}{1-a^2}+\frac{125}{84}\cdot\frac{2a}{1+a^2}+\frac{4625}{1008}\cdot\frac{1-a^2}{1+a^2}\leq\frac{1339}{252},$$ $$\frac{2b}{1-b^2}+\frac{125}{84}\cdot\frac{2b}{1+b^2}+\frac{4625}{1008}\cdot\frac{1-b^2}{1+b^2}\leq\frac{1339}{252}$$ and $$\frac{2c}{1-c^2}+\frac{125}{84}\cdot\frac{2c}{1+c^2}+\frac{4625}{1008}\cdot\frac{1-c^2}{1+c^2}\leq\frac{591}{112}.$$ The first inequality gives: $$\frac{(3a-1)^2(731-630a-1109a^2)}{1-a^4}\geq0$$ and the third gives $$\frac{(2c-1)^2(347-1120c-1243c^2)}{1-c^4}\geq0.$$

Answer 2

A proof following user51547 and Maestro13's nice idea.

WLOG, assume that $x, y, z\in [0, 2\pi]$. Clearly, $\cos x, \cos y, \cos z, \sin x, \sin y, \sin z > 0$. Thus, $x, y, z \in (0, \pi/2)$.

We claim that $x + y + z > \pi/2$. Indeed, if $x + y + z \le \pi/2$, we have $x + y \le \pi/3$ and $\sin x + \sin y = 2\sin\frac{x + y}{2}\cos \frac{x - y}{2} \le 1$ which contradicts $\sin x + \sin y + \sin z = 2$.

Also, we can prove that $x + y + z \le \pi - \arcsin\frac45$. See this question.

Now let us express $\tan x + \tan y + \tan z$ in terms of $abc$.

Let $a = \cos x + \mathrm{i}\sin x, b = \cos y + \mathrm{i}\sin y$, and $c = \cos z + \mathrm{i} \sin z$.

We have $$a + b + c = \frac{11}{5} + 2\mathrm{i},\quad \frac{1}{a} + \frac{1}{b} + \frac{1}{c} = \frac{11}{5} - 2\mathrm{i}. \tag{1}$$

Let $abc = u + \mathrm{i}v$. We have $$u = \cos(x + y + z), \quad v = \sin(x + y + z), \quad \frac{1}{abc} = u - \mathrm{i}v. \tag{2}$$

Using $\frac{a - 1/a}{a + 1/a} = \mathrm{i} \tan x$ etc., we have \begin{align*} &\mathrm{i}(\tan x + \tan y + \tan z)\\[6pt] ={}& \frac{a - 1/a}{a + 1/a} + \frac{b - 1/b}{b + 1/b} + \frac{c - 1/c}{c + 1/c}\\[6pt] ={}& \frac{3(abc - \frac{1}{abc}) + (\frac{bc}{a} + \frac{ca}{b} + \frac{ab}{c}) - (\frac{a}{bc} + \frac{b}{ca} + \frac{c}{ab})}{(abc + \frac{1}{abc}) + (\frac{bc}{a} + \frac{ca}{b} + \frac{ab}{c}) + (\frac{a}{bc} + \frac{b}{ca} + \frac{c}{ab})}\\[6pt] ={}& \frac{6v\mathrm{i} + [(\frac{11}{5} - 2\mathrm{i})^2(u + \mathrm{i} v) - 2(\frac{11}{5} + 2\mathrm{i})] - [(\frac{11}{5} + 2\mathrm{i})^2(u - \mathrm{i} v) - 2(\frac{11}{5} - 2\mathrm{i})]}{2u + [(\frac{11}{5} - 2\mathrm{i})^2(u + \mathrm{i} v) - 2(\frac{11}{5} + 2\mathrm{i})] + [(\frac{11}{5} + 2\mathrm{i})^2(u - \mathrm{i} v) - 2(\frac{11}{5} - 2\mathrm{i})]}\\[6pt] ={}& \mathrm{i}\cdot \frac{-110u + 48v - 50}{23u + 110v - 55} \tag{3} \end{align*} where we use (1), (2), and \begin{align*} \frac{(a + b + c)^2}{abc} &= \frac{a}{bc} + \frac{b}{ca} + \frac{c}{ab} + 2\left(\frac{1}{a} + \frac{1}{b} + \frac{1}{c}\right), \\[6pt] \left(\frac{1}{a} + \frac{1}{b} + \frac{1}{c}\right)^2abc &= \frac{bc}{a} + \frac{ca}{b} + \frac{ab}{c} + 2(a + b + c). \end{align*}

From (3), we have $$\tan x + \tan y + \tan z = \frac{-110u + 48v - 50}{23u + 110v - 55} = \frac{-110\cos w + 48\sin w - 50}{23\cos w + 110\sin w - 55}$$ where $w := x + y + z \in (\pi/2, \pi - \arcsin\frac45]$.

It suffices to prove that, for all $w\in (\pi/2, \pi - \arcsin\frac45]$, $$\frac{-110\cos w + 48\sin w - 50}{23\cos w + 110\sin w - 55} \le \frac{17}{6}.$$

Let $$f(w) := \frac{-110\cos w + 48\sin w - 50}{23\cos w + 110\sin w - 55}.$$

We have $$f'(w) = \frac{13204 - 7200\sin w + 2860\cos w}{(23\cos w + 110\sin w - 55)^2} > 0$$ where we use $13204 - \sqrt{7200^2 + 2860^2} > 0$.

Also, we have $f(\pi - \arcsin\frac45) = \frac{17}{6}$. Thus, we have $f(w) \le \frac{17}{6}$ for all $w \in (\pi/2, \pi - \arcsin\frac45]$.

We are done.