Minimum value of $a^2 \cot (10^ \circ)+b^2\cot (70^ \circ)+c^2\cot (130^ \circ)$ Let $a,b,c$ be real numbers such that $a+b+c=3$, then find minimum value of
$$a^2 \cot (10^ \circ)+b^2\cot (70^ \circ)+c^2\cot (130^ \circ)$$
I have solved many such question using A.M.-G.M. inequality but since $\cot (130^ \circ)$ is negative, hence it cannot be applied here. I also have a result in mind which is
$$\cot (60 ^\circ-\theta)\cot (\theta)\cot (60 ^\circ+\theta)=\cot (3\theta)$$ and in this question if we write $
\cot(130^\circ)=-\cot (50^\circ)$, then above mentioned result can be applied on $\cot (10^ \circ),\cot (70^ \circ),\cot (50^ \circ)$ but I am not able to put everything together and reach the final answer which is given as $\sqrt{27}$. Any help or hint would be appreciated.

Since $\cot (10^ \circ)$ and $\cot (70^ \circ)$ are both positive and $a^2$ and $b^2$ are positive, because they are real, you want to minimize their contribution. Set $a=b=0$, then $c=3$ and the minimum value is $9\cot (130^ \circ) < 0$ so the answer is wrong unless we take the negative root.
$$\sqrt{27} = -3 \sqrt{3} $$ but we know that
$$\cot (120^ \circ)=\frac{-\sqrt{3}}{3}$$ so $$9 \cot (120^ \circ)=-3 \sqrt{3}=\sqrt{27}$$ so I think your book has a misprint.
