let $\Delta ABC,\Delta XYZ$ are acute triangle
show that $$\max\lbrace\cot{A}(\cot{Y}+\cot{Z}),\cot{B}(\cot{Z}+\cot{X}),\cot{C}(\cot{X}+\cot{Y})\rbrace\ge\dfrac{2}{3}$$
My idea: since $$\cot{Y}+\cot{Z}=\dfrac{\cos{Y}}{\sin{Y}}+\dfrac{\cos{Z}}{\sin{Z}}=\dfrac{\sin{Z}\cos{Y}+\sin{Y}\cos{Z}}{\sin{Y}\sin{Z}}=\dfrac{\sin{(Y+Z)}}{\sin{Y}\sin{Z}}=\dfrac{\sin{X}}{\sin{Y}\sin{Z}}$$ so $$a=\cot{A}(\cot{Y}+\cot{Z})=\cot{A}\dfrac{\sin{X}}{\sin{Y}\sin{Z}}$$ and the simaler we have $$b=\cot{B}(\cot{Y}+\cot{Z})=\dfrac{\cot{C}\sin{Y}}{\sin{X}\sin{Z}},c=\cot{C}(\cot{X}+\cot{Y})=\cot{C}\dfrac{\sin{Z}}{\sin{X}\sin{Y}}$$ so $$abc=\dfrac{cot{A}\cot{B}\cot{C}}{\sin{X}\sin{Y}\sin{Z}}$$
since we kow $$\cot{A}\cot{B}\cot{C}\ge\sqrt{3},\sin{X}\sin{Y}\sin{Z}\le\dfrac{3\sqrt{3}}{8}$$ so $$abc\ge \dfrac{8}{3}$$
so $$\max\lbrace a,b,c\rbrace\ge\dfrac{2\sqrt{3}}{3}\ge\dfrac{2}{3}$$
My methods is true?
This problem is also from China South East Mathematical Olympiad
