Prove that $\tan A+\tan B+\tan C\leq6\sqrt3-9$, where $A,B$ and $C\in(0,\pi/2)$ and $A+B+C=\pi$.

I tried using Jensen's Inequality, but it only gives a lower bound since $\tan x$ is a convex function over the interval $(0,\pi/2)$. Moreover, I tried using $QM\geq AM,$ but that still gave me a convex function $f(x)=\sec^2(x)$ which made it impossible for me to obtain an upper bound.

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    $\begingroup$ Unless I'm missing something: $\tan \frac{\pi}{3} + \tan \frac{\pi}{3} + \tan \frac{\pi}{3} = 3 \sqrt{3} \approx 5.19615 > 1.3923 \approx 6\sqrt{3} - 9$ $\endgroup$ – Sirzh Oct 26 '16 at 11:14
  • $\begingroup$ What makes you think there should be an upper bound $\endgroup$ – David Quinn Oct 26 '16 at 11:19
  • $\begingroup$ If $A=B=\frac{\pi}{2}-\epsilon$ and $C=2\epsilon$ for $0<\epsilon<\frac{\pi}{2}$, the sum is greater than $2\tan(\frac{\pi}{2}-\epsilon)$ which is clearly unbounded. Maybe it's a lower bound you're looking for ? $\endgroup$ – Nicolas FRANCOIS Oct 26 '16 at 11:49

$f(x)=\tan x $ is convex function when $x\in(0,\pi/2).$ $$$$ Therefore, $$f(\frac{A+B+C}{3})\le\frac{f(A)+f(B)+f(C)}{3}$$ Therefore$$\tan A+ tan B+ \tan C\geq3\tan(\frac{(A+B+C}{3})=3\sqrt{3}$$obvious don't have the maximum.


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