If $x>0, y>0,x+y=\frac{\pi}{3}$ then maximum value of $\tan x\tan y$ 
If $x>0, y>0$ and $x+y=\frac{\pi}{3}$, then find the maximum value of $\tan x\tan y$

My Attempt
$x>0, y>0, x+y=\frac{\pi}{3}\implies x, y$ in $1^\text{st}$ quadrant.
$\tan x, \tan y>0, \tan(x+y)=\sqrt{3}, \tan x\tan y>0$
$$
\tan x+\tan y\geq2\sqrt{\tan x.\tan y}\implies \tan^2x+\tan^2y+2\tan x\tan y\geq 4\tan x\tan y\\
2\tan x\tan y\leq\tan^2x+\tan^2y\implies \color{red}{?}
$$
or
$$
1-\tan x\tan y=\frac{\tan x+\tan y}{\tan(x+y)}\implies\tan x\tan y=1-\frac{\tan x+\frac{\sqrt{3}-\tan x}{1+\sqrt{3}\tan x}}{\sqrt{3}}\\
\tan x\tan y=1-\frac{\frac{\tan x+\sqrt{3}\tan^2x+\sqrt{3}-\tan x}{1+\sqrt{3}\tan x}}{\sqrt{3}}=1-\frac{1+\tan^2x}{1+\sqrt{3}\tan x}\leq\color{red}{?}\\
=\frac{1+\sqrt{3}\tan x-1-\tan^2x}{1+\sqrt{3}\tan x}=\frac{\tan x(\sqrt{3}-\tan x)}{1+\sqrt{3}\tan x}=\color{red}{?}
$$
Note: I prefer not to do differentiation
 A: Hint: One way is finding the extrimum of $f(x,y)=\tan x\tan y$ with condition $x+y=\dfrac{\pi}{3}$, then with
$$F(x,y)=\tan x\tan y-\lambda\left(x+y-\dfrac{\pi}{3}\right)$$
we have 
$$\tan y=\lambda\cos^2x~~~,~~~\tan x=\lambda\cos^2y$$
then
$$\tan y\lambda\cos^2y=\tan x\lambda\cos^2x$$
shows $\sin x\cos x=\sin y\cos y$ or $\color{blue}{x=y}$.
A: $$F(x)=\tan x\tan y=\dfrac{2\sin x\sin y}{2\cos x\cos  y}=\dfrac{\cos(x-y)-\cos(x+y)}{\cos(x-y)+\cos(x+y)}=1-\dfrac{2\cos(x+y)}{\cos(x-y)+\cos(x+y)}$$
Method$\#1:$
For $x+y=\dfrac\pi3$
$$F(x)=1-\dfrac2{2\cos\left(2x-\dfrac\pi3\right)+1}$$ which will be maximum 
if $-\dfrac2{2\cos\left(2x-\dfrac\pi3\right)+1}$ is maximum
if $\dfrac2{2\cos\left(2x-\dfrac\pi3\right)+1}$ is minimum positive
if $2\cos\left(2x-\dfrac\pi3\right)+1$ is maximum positive
Method$\#2:$
For $x+y=\dfrac\pi3,$
$F(x)=\dfrac{2\cos\left(2x-\dfrac\pi3\right)-1}{2\cos\left(2x-\dfrac\pi3\right)+1}$ 
$\iff2\cos\left(2x-\dfrac\pi3\right)=\dfrac{1+F(x)}{1-F(x)}$
$0<x<\dfrac\pi3,2\ge2\cos\left(2x-\dfrac\pi3\right)\ge\dfrac12$
$2\ge\dfrac{1+F(x)}{1-F(x)}\ge\dfrac12$
$2\ge\dfrac{1+F(x)}{1-F(x)}\iff\dfrac{1-3F(x)}{1-F(x)}\ge0\implies$ either $F(x)\le\dfrac13$ or $F(x)>1$
$\dfrac{1+F(x)}{1-F(x)}\ge\dfrac12\iff\dfrac{1+3F(x)}{1-F(x)}\ge0\iff\dfrac{F(x)+\dfrac13}{F(x)-1}\le0\iff-\dfrac13\le F(x)<1$
$\implies F(x)\le\dfrac13$ which occurs if $\cos\left(2x-\dfrac\pi3\right)=1$
A: For $-\dfrac{\pi}{6} < t < \dfrac{\pi}{6}$, let $x = \dfrac{\pi}{6} - t \ $ and $ \ y = \dfrac{\pi}{6} + t$.
Then $x > 0$, $y>0$ and $x + y = \dfrac{\pi}{3}$.
Let
\begin{align}
   f(t) 
   &= \tan(x) \tan(y) \\
   &= \tan\left(\dfrac{\pi}{6} - t \right) \tan \left(\dfrac{\pi}{6} + t \right) \\
   &= \dfrac{1-\sqrt 3 \tan t}{\sqrt 3 + \tan t} \cdot
      \dfrac{1+\sqrt 3 \tan t}{\sqrt 3 - \tan t} \\
   &= \dfrac{1 - 3 \tan^2 t}{3 - \tan^2 t} \\
   &= \dfrac{3 \tan^2 t - 1}{\tan^2 t - 3} \\
   &= 3 - \dfrac{8}{3 - \tan^2 t}
\end{align}
Note that $f(t)$ is an even function on the interval $\left[-\dfrac{\pi}{6}, \dfrac{\pi}{6} \right]$. 
\begin{align}
   t \in \left[-\dfrac{\pi}{6}, 0 \right]
   &\implies -\dfrac{1}{\sqrt 3} \le \tan t \le 0 \\
   &\implies 0 \le \tan^2 t \le \dfrac 13 \\
   &\implies \dfrac 83 \le 3 - \tan^2 t \le 3 \\
   &\implies  \dfrac 83 \le \dfrac{8}{3 - \tan^2 t} \le 3 \\
   &\implies 0 \le f(t) \le \dfrac 13
\end{align}
