Finding the minimum value of $\cot^2A + \cot^2B+ \cot^2C$ where $A$, $B$ and $C$ are angles of a triangle.

Question

The question is:

If $A+B+C= \pi$, where $A>0$, $B>0$, $C>0$, then find the minimum value of $$\cot^2A+\cot^2B +\cot^2C.$$

My solution:

$(\cot A + \cot B + \cot C)^2\ge0$ // square of a real number

$\implies \cot^2A +\cot^2B + \cot^2 +2 \ge0 $ //Conditional identity used: $\cot A \cot B + \cot B \cot C + \cot A \cot C =1$

$\implies \cot^2A +\cot^2B + \cot^2 C \ge -2$

Thus according to me the answer should be $-2$. However, the answer key states that the answer is $1$. Where have I gone wrong?

Answer 1

Since $\cot^2$ is a convex function, we can use Jensen. $$\sum_{cyc}\cot^2\alpha\geq3\cot^2\frac{\alpha+\beta+\gamma}{3}=1.$$ The equality occurs for $\alpha=\beta=\gamma=\frac{\pi}{3},$ which says that $1$ is a minimal value.

Actually, $(\cot^2x)''=\frac{2(2+\cos2x)}{\sin^4x}>0$.

Also, we can use the following way.

Let $a$, $b$ and $c$ be sides-lengths and $S$ be an area of the triangle.

Thus, we need to prove that $$\sum_{cyc}\frac{\cos^2\alpha}{\sin^2\alpha}\geq1$$ or $$\sum_{cyc}\frac{\frac{(b^2+c^2-a^2)^2}{4b^2c^2}}{\frac{4S^2}{b^2c^2}}\geq1$$ or $$\sum_{cyc}(b^2+c^2-a^2)^2\geq16S^2$$ or $$\sum_{cyc}(b^2+c^2-a^2)^2\geq\sum_{cyc}(2a^2b^2-a^4)$$ or $$\sum_{cyc}(a^4-a^2b^2)\geq0$$ or $$\sum_{cyc}(a^2-b^2)^2\geq0.$$ Done!

Also we can make the following. It's what you wish. I think. $$\sum_{cyc}\cot^2\alpha=1+\sum_{cyc}(\cot^2\alpha-\cot\alpha\cot\beta)=1+\frac{1}{2}\sum_{cyc}(\cot\alpha-\cot\beta)^2\geq1$$

It's $$\cot^2\alpha+\cot^2\beta+\cos^2\gamma=$$ $$=1+\cot^2\alpha+\cot^2\beta+\cos^2\gamma-\cot\alpha\cot\beta-\cot\alpha\cot\gamma-\cot\beta\cot\gamma=$$ $$=1+$$ $$+\frac{1}{2}\left(\cot^2\alpha-2\cot\alpha\cot\beta+\cot^2\beta+\cot^2\alpha-2\cot\alpha\cot\gamma+\cos^2\gamma+\cot^2\beta-2\cot\beta\cot\gamma+\cot^2\gamma\right)=$$ $$=1+\frac{1}{2}\left((\cot\alpha-\cot\beta)^2+(\cot\alpha-\cot\gamma)^2+(\cot\beta-\cot\gamma)^2\right)\geq1.$$

Answer 2

it is equivalent to $$\frac{1}{\sin(A)^2}+\frac{1}{\sin(B)^2}+\frac{1}{\sin(C)^2}\geq 4$$ with $$\sin(A)=\frac{a}{2R}$$ etc and $$S=\sqrt{s(s-a)(s-b)(s-c)}$$ and $$S=\frac{abc}{4R}$$ we get $$b^2c^2+c^2a^2+a^2b^2-(-a+b+c)(a-b+c)(a+b-c)(a+b+c)\geq 0$$ and this is equivalent to $$a^4+b^4+c^4\geq a^2b^2+b^2c^2+c^2a^2$$ which is true.

Answer 3

Using Prove that $\tan A + \tan B + \tan C = \tan A\tan B\tan C,$ $A+B+C = 180^\circ$,

$$\cot A\cot B+\cot B\cot C+\cot C\cot A=1$$

Now $$(\cot A-\cot B)^2+(\cot B-\cot C)^2+(\cot C-\cot A)^2\ge0$$

$$\iff\cot^2A+\cot^2B+\cot^2C\ge\cot A\cot B+\cot B\cot C+\cot C\cot A$$