If $A,B>0$ and $A+B = \frac{\pi}{3},$ Then find Maximum value of $\tan A\cdot \tan B$. If $A,B>0$ and $\displaystyle A+B = \frac{\pi}{3},$ Then find Maximum value of $\tan A\cdot \tan B$.
$\bf{My\; Try::}$ Given $$\displaystyle A+ B = \frac{\pi}{3}$$ and $A,B>0$.
So we can say $$\displaystyle 0< A,B<\frac{\pi}{3}$$. Now taking $\tan $ on both side, we get
$$\displaystyle \tan(A+B) = \tan \left(\frac{\pi}{3}\right).$$ So $\displaystyle \frac{\tan A+\tan B}{1-\tan A\tan B} = \sqrt{3}$.
Now Let $\displaystyle \tan A\cdot \tan B=y\;,$ Then $\displaystyle \tan B = \frac{y}{\tan A}.$
So $$\displaystyle \frac{\tan A+\frac{y}{\tan A}}{1-y}=\sqrt{3}\Rightarrow \tan^2 A+y=\sqrt{3}\tan A-y\sqrt{3}\tan A$$
So equation $$\tan^2 A+\sqrt{3}\left(y-1\right)\tan A+y=0$$
Now for real values of $y\;,$ Given equation has real roots. So its $\bf{Discrimnant>0}$
So $$\displaystyle 3\left(y-1\right)^2-4y\geq 0\Rightarrow 3y^2+3-6y-4y\geq 0$$
So we get $$3y^2-10y+3\geq 0\Rightarrow \displaystyle 3y^2-9y-y+3\geq 0$$
So we get $$\displaystyle y\leq \frac{1}{3}\cup y\geq 3$$, But above we get $\displaystyle 0<A,B<\frac{\pi}{3}$
So We Get $$\bf{\displaystyle y_{Max.} = \left(\tan A \cdot \tan B\right)_{Max} = \frac{1}{3}}.$$
My Question is can we solve above question using $\bf{A.M\geq G.M}$ Inequality or Power Mean equality.
Thanks 

By using $1-\tan A\tan B=\frac{\tan A+\tan B}{\tan(A+B)}$, it follows that for $A>0,B>0,A+B=\pi/3$,
$$\tan A\tan B=\frac{\tan A(\sqrt{3}-\tan A)}{1+\sqrt{3}\tan A}.$$
Hence
$$\begin{align}\max_{A>0,B>0,A+B=\pi/3} \tan A\tan B&=\max_{0<A<\pi/3} \frac{\tan A(\sqrt{3}-\tan A)}{1+\sqrt{3}\tan A}=\max_{1<s<4} \frac{(s-1)(4-s)}{3s}\\
&=\frac{1}{3}\max_{1<s<4}\left(5-\left(s+\frac{4}{s}\right)\right)=
\frac{1}{3}\left(5-\min_{1<s<4}\left(s+\frac{4}{s}\right)\right)
\\&=\frac{1}{3}(5-4)=\frac{1}{3}\end{align}$$
where $s=1+\sqrt{3}\tan A$ and by AGM-inequality
$$s+\frac{4}{s}\geq 2\sqrt{s\cdot \frac{4}{s}}=4$$
with equality for $s=2\in(1,4)$, that is $A=B=\pi/6$.
