Baltic Way 1997 · Problem 14

Denote $|B C|=a,|C A|=b$ and $|A B|=c$, then we have $2 b^{2}=a^{2}+c^{2}$. Applying the cosine and sine laws to triangle $A B C$ we have:

$$\begin{aligned} & \cot B=\frac{\cos B}{\sin B}=\frac{\left(a^{2}+c^{2}-b^{2}\right) \cdot 2 R}{2 a c \cdot b}=\frac{\left(a^{2}+c^{2}-b^{2}\right) \cdot R}{a b c}, \\ & \cot A=\frac{\cos A}{\sin A}=\frac{\left(b^{2}+c^{2}-a^{2}\right) \cdot R}{a b c}, \end{aligned}$$ $$\cot C=\frac{\cos C}{\sin C}=\frac{\left(a^{2}+b^{2}-c^{2}\right) \cdot R}{a b c}$$

where $R$ is the circumradius of triangle $A B C$. To finish the proof it hence suffices to show that $\left(a^{2}+c^{2}-b^{2}\right)^{2} \geqslant\left(b^{2}+c^{2}-a^{2}\right)\left(a^{2}+b^{2}-c^{2}\right)$. Indeed, from the AM-GM inequality we get

$$\begin{aligned} \left(b^{2}+c^{2}-a^{2}\right)\left(a^{2}+b^{2}-c^{2}\right) & \leqslant \frac{\left(b^{2}+c^{2}-a^{2}+a^{2}+b^{2}-c^{2}\right)^{2}}{4}=b^{4}= \\ & =\left(2 b^{2}-b^{2}\right)^{2}=\left(a^{2}+c^{2}-b^{2}\right)^{2} . \end{aligned}$$

Figure 9