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This question is picked from AM GM HM inequalities, so this is to be proved form that concept only, I think it isn't possible because there is no inequality, but if it is please tell me how.

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For acute triangles the relation $\tan(A)+\tan(B)+\tan(C)=\tan(A)\tan(B)\tan(C)$ can be combined with the AM-GM inequality. If $s$ is the sum of tangents and $p$ the product one gets $(s/3)^3 \geq p = s$ so that $s^2 \geq 3^3$ with equality for equilateral triangles.

For obtuse triangles the sum of tangents is negative, and for right triangles the sum is infinite. In neither case can it equal $3 \sqrt{3}$.

More simply, for acute angles the tangent function is convex, so $\tan(A)+\tan(B)+\tan(C) \geq \tan(\frac{A+B+C}{3}) = 3 \sqrt{3}$.

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