Question

cotA+cotB+cotC\(\ge\)\(\sqrt{3}\)

cm bất đẳng thức

Answer 1

Xét trong 1 tam giác:

\(\tan A+\tan B+\tan C=\tan\left(A+B\right).\left(1-\tan A.\tan B\right)+\tan C\)

\(=\tan\left(\pi-C\right)\left(1-\tan A.\tan B\right)+\tan C\)

\(=\tan A.\tan B.\tan C\)

☕ Quay lại bài toán, cần chứng minh \(\dfrac{1}{\tan A}+\dfrac{1}{\tan B}+\dfrac{1}{\tan C}\ge\sqrt{3}\)

Theo AM-GM:

\(VT^2\ge3\left(\dfrac{1}{\tan A.\tan B}+\dfrac{1}{\tan B.\tan C}+\dfrac{1}{\tan C.\tan A}\right)\)

\(=\dfrac{3\left(\tan A+\tan B+\tan C\right)}{\tan A.\tan B.\tan C}=3\). Suy ra đpcm