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.
Best Answer
There's neither a maximum nor a minimum value. Take $a=1,b=x\in\mathbb{R},c=2-x$. Then $$a^2\cot(10^\circ)+b^2\cot(70^\circ)+c^2\cot(130^\circ)\\ =(\cot(130^\circ)+\cot(70^\circ))x^2-4\cot(130^\circ)x+4\cot(130^\circ)+\cot(10^\circ)\xrightarrow{x\to\infty}-\infty $$ since the above is a quadratic function with negative leading coefficient: $\cot(130^\circ)+\cot(70^\circ)=\cot(70^\circ)-\cot(50^\circ)<0$. So the expression is unbounded below. Similarly, taking $a=2-x,b=x,c=1$ leads to $$a^2\cot(10^\circ)+b^2\cot(70^\circ)+c^2\cot(130^\circ)\\ =(\cot(10^\circ)+\cot(70^\circ))x^2-4\cot(10^\circ)x+4\cot(10^\circ)+\cot(130^\circ)\xrightarrow{x\to\infty}\infty $$ since $\cot(10^\circ)+\cot(70^\circ)>0$. Thus, the expression is also unbounded above.
Where does the $\sqrt{27}$ you mentioned come from? If we set up Lagrange multipliers for this problem (the calculations here are relatively straightforward), we obtain the point $(a,b,c)=\left(-\cot(70^{\circ})\cot(130^{\circ}),-\cot(10^{\circ})\cot(130^{\circ}),-\cot(10^{\circ})\cot(70^{\circ})\right)$ and the expression is equal to $\sqrt{27}$ at this point. However, Lagrange multipliers do not always give us the extrema. In this particular case, we have a saddle point there.