Minimizing $\cot^2 A +\cot^2 B + \cot^2 C$ for $A+B+C=\pi$

Question

If $A + B + C = \pi$, then find the minimum value of $\cot^2 A +\cot^2 B + \cot^2 C$.

I don't know how to solve it. And can you please mention the used formulas first.

What I can see is that if one of the angles $A$, $B$, $C$ is small, then the value $\cot^2A$ or $\cot^2B$ or $\cot^2C$ will be big. So I want to make angles big (more precisely, close to $\pi/2$, where cotangent is zero), but the condition $A+B+C=\pi$ prevents me from making all three of them very big .

I can see that if $A=B=C=\frac\pi3$, then I get $\cot A=\cot B=\cot C=\frac1{\sqrt3}$ and $\cot^2A+\cot^2B+\cot^2C=1$. But I do not know whether this is indeed minimum. (According to WolframAlpha this is the minimum. However, I would like to see some proof of this fact.)

Answer 1

Put $x=\cot A,y=\cot B$. Using the standard formulae we have $z=\cot C=-\cot(A+B)=\frac{1-xy}{x+y}$. So we want to minimize $$f(x,y)=\frac{(x^2+y^2)(x+y)^2+(xy-1)^2}{(x+y)^2}$$ where $x,y$ can take any real values.

If we want to minimize $(x^2+y^2)(x+y)^2+x^2y^2-2xy+1$ subject to $x+y=k$, then using Lagrange multipliers we find $4x^3-2y+6x^2y+6xy^2+2y^3=\lambda=-2x+2x^3+6yx^2+6xy^2+4y^3$ and so $x=y$ or $x^2+xy+y^2=0$. Note that this result is independent of $k$. Note also that $f(x,y)$ is large for $|x|$ or $|y|$ large, so the minimum will occur at a stationary point.

If $x=y$ then we want to minimize $2x^2+\frac{(x^2-1)^2}{4x^2}{4x^2}$. Differentiating, this has a minimum at $x=\frac{1}{\sqrt3}$ that makes $A=B=C=\frac{\pi}{3}$ and the minimum value 1.

If $x^2+xy+y^2=0$, then $-xy=x^2+y^2\ge0$. But $0\le(x+y)^2=x^2+xy+y^2+xy=xy$, so we must have $xy=0$ and hence also $x^2+y^2=0$ and so $x=y=0$, which clearly does not yield a minimum.

----------Added later, courtesy of Martin Sleziak ---------

I am not particularly happy with the argument above, which seems like using a hammer to crack a nut. The OP in this earlier question In a $\triangle ABC,$ Evaluation of minimum value of $\cot^2 A+\cot^2 B+\cot^2 C$ provided a much simpler argument:

By AM/GM we have $\cot^2A+\cot^2B\ge 2\cot A\cot B$, and similarly $\cot^2B+\cot^2C\ge 2\cot B\cot C$ and $\cot^2C+\cot^2A\ge 2\cot C\cot A$ so we have $$\cot^2A+\cot^2B+\cot^2C\ge \cot A\cot B+\cot B\cot C+\cot C\cot A$$ But now by the formula at the start we have $\cot C(\cot A+\cot B)=1-\cot A\cot B$ which gives immediately $$\cot^2A+\cot^2B+\cot^2C\ge1$$ Instead of using AM/GM we could also use the rearrangement inequality.

If you like that argument upvote his question rather than this answer!

Answer 2

Since the cotangent function is odd, we may assume $A,B,C\geq 0$ without loss of generality.
So $A,B,C$ are the angles of a triangle. Let we deal first with the case of an acute triangle.
By the Cauchy-Schwarz inequality:

$$\cot(A)^2+\cot(B)^2+\cot(C)^2 \geq 3\left(\frac{\cot A+\cot B+\cot C}{3}\right)^2 = \frac{\cot(\omega)^2}{3}$$ where $\omega$ is the Brocard angle. Since $\omega\leq\frac{\pi}{6}$,

$$ \cot(A)^2+\cot(B)^2+\cot(C)^2 \geq \color{red}{1}$$

and equality is attained by the equilateral triangle. Now we may deal with the obtuse case, and assume $A\geq \frac{\pi}{2}$. So $B,C$ are acute angles and their sum does not exceed $\frac{\pi}{2}$. Since the $\cot^2$ function is convex and decreasing on the interval $\left(0,\frac{\pi}{2}\right)$, we have: $$ \cot^2(B)+\cot^2(C) \geq 2\cot^2\left(\frac{B+C}{2}\right)\geq 2\cot^2\left(\frac{\pi}{4}\right)=2 $$ and we are done.

Answer 3

For another way, since $\cot^2x$ is convex when $x \in [0,\pi]$ and we need to worry only of positive $x$, so by Jensen's inequality: $$ \sum_{cyc} \cot^2 A \geqslant 3 \cot^2\frac{A+B+C}3=1$$

As equality is possible when $A=B=C=\frac{\pi}3$, we have the minimum.