$\dfrac{\tan x}{1+\tan y.\tan z}+\dfrac{\tan y}{1+\tan z.\tan t}+\dfrac{\tan z}{1+\tan t.\tan x}+\dfrac{\tan t}{1+\tan x.\tan y}=2$

Bài toán:
Tìm các góc $x,y,z,t$ biết
$\left\{\begin{matrix}0<x,y,z,t<\dfrac{\pi}{2},\tan x+\tan y+\tan z+\tan t=4  &  & \\ \dfrac{\tan x}{1+\tan y.\tan z}+\dfrac{\tan y}{1+\tan z.\tan t}+\dfrac{\tan z}{1+\tan t.\tan x}+\dfrac{\tan t}{1+\tan x.\tan y}=2  &  &  \end{matrix}\right.$.

(Đề thi cuối học kì 1 lớp 11 trường THPT Nguyễn Du, Thái Bình, năm học 2015-2016)

Lời giải:

Đặt $\left\{\begin{matrix}\tan x=a  &  & \\ \tan y=b  &  & \\ \tan z=c \\ \tan t=d \end{matrix}\right.(a,b,c,d>0)\Rightarrow a+b+c+d=4$.

$PT(2)\Leftrightarrow \sum \dfrac{a}{1+bc}=2$.

Có: 

$\begin{align*} \sum \dfrac{a}{1+bc}&=\sum \dfrac{a^2}{a+abc}\\ &\ge \dfrac{(a+b+c+d)^2}{a+b+c+d+abc+bcd+cda+dab}\\ &=\dfrac{16}{4+abc+bcd+cda+dab}~(*) \end{align*}$

Ta có:
$\begin{align*} abc+bcd+cda+dab&=ab(c+d)+cd(a+b)\\ &\le \dfrac{(a+b)^2}{4}(c+d)+\dfrac{(c+d)^2}{4}(a+b)\\ &=(a+b)(c+d).\dfrac{a+b+c+d}{4}\\ &\le \dfrac{(a+b+c+d)^2}{4}.\dfrac{a+b+c+d}{4}\\ &=4 \end{align*}$

Khi đó: $(*)\Leftrightarrow \sum \dfrac{a}{1+bc}\ge \dfrac{16}{4+4}=2$

Mà $\sum \dfrac{a}{1+bc}=2$.

Dấu "=" xảy ra khi: $a=b=c=d=1$ hay $\tan x=\tan y=\tan z=\tan t=1\Rightarrow x=y=z=t=\dfrac{\pi}{4}$.