<s>[INST] 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
 [/INST] 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 $\color{blue}{x=y}$.
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