<s>[INST] Prove that $\frac{2}{1+\tan A}+\frac{2}{1+\tan B}+\frac{2}{1+\tan C} \le 3(\sqrt{3}-1)$ let $ABC$ be an acute triangle with all angles greater than $45^o$
Prove that
$$\frac{2}{1+\tan A}+\frac{2}{1+\tan B}+\frac{2}{1+\tan C} \le 3(\sqrt{3}-1)$$
I let $\tan A=a$, $\tan B=b$, $\tan C=c$ with $a+b+c=abc$ then the inequality  equivalent to $\frac{2}{1+a}+\frac{2}{1+b}+\frac{2}{1+c} \le 3(\sqrt{3}-1)$ but it's not work.
I don't known what to do. Algebra or trigonometric?
 [/INST] Let's use Lagrange multiplier method to maximize 
$$
    f(a,b,c) = \frac{2}{1+a} + \frac{2}{1+b} + \frac{2}{1+c}
$$
subject to the constraint $0 = g(a,b,c) = a+b+c-a b c$. Derivatives of $f - \lambda g$ read:
$$ \begin{eqnarray}
  \frac{\partial(f- \lambda g)}{\partial a} &=& -\frac{2}{(1+a)^2} + \lambda (b c -1) = \frac{-2+ \lambda (b c-1)(1+a)^2}{(1+a)^2} = 0\\
  \frac{\partial(f- \lambda g)}{\partial b} &=& -\frac{2}{(1+b)^2} + \lambda (a c -1) = \frac{-2+ \lambda (a c-1)(1+b)^2}{(1+b)^2} = 0\\
 \frac{\partial(f- \lambda g)}{\partial c} &=& -\frac{2}{(1+c)^2} + \lambda (a b  -1) = \frac{-2+ \lambda (a b-1)(1+c)^2}{(1+c)^2} = 0
\end{eqnarray}
$$
Eliminating out $\lambda$ this results into two equation:
$$
    (1+a)^2 (b c-1) = (1+b)^2 (a c-1) , \quad (1+b)^2 (a c-1) = (1+c)^2 (a b-1)
$$
substituting $c = \frac{a+b}{a b-1}$ we get:
$$
   \frac{(1+a)^2 (1+b^2}{a b -1} = \frac{(1+a^2)(1+b)^2}{a b -1}, \quad 
   \frac{(1+a^2)(1+b)^2}{a b -1} = \frac{\left(a b + a +b -1\right)^2}{a b -1}
$$
Subtracting:
$$
    \frac{2 (b-a)(a b -1)}{a b-1} = 0, \quad \frac{4 b - 2 a(b^2-1)}{a b -1} = 0
$$
Giving $a=b$ and $a^2 = 3$. Thus $a=b=c=\sqrt{3}$ is the extremum. It is easy to see that this is a maximum. The inequality, we see, is saturated at $a=b=c=\sqrt{3}$, corresponding to $A=B=C=\frac{\pi}{3}$.
</s>