Xét ΔABC có

\(cosA=\frac{b^2+c^2-a^2}{2bc};cosB=\frac{a^2+c^2-b^2}{2ac};cosC=\frac{a^2+b^2-c^2}{2ab}\)

Xét ΔABC có \(S_{ABC}=\frac12\cdot bc\cdot\sin A=\frac12\cdot ac\cdot\sin B=\frac12\cdot ab\cdot\sin C\)

=>\(\sin A=\frac{2\cdot S}{bc};\sin B=\frac{2\cdot S}{ac};\sin C=\frac{2\cdot S}{ab}\)

\(\cot A=\frac{cosA}{\sin A}=\frac{b^2+c^2-a^2}{2bc}:\frac{2S}{bc}=\frac{b^2+c^2-a^2}{4S}\)

\(\cot B=\frac{cosB}{sinB}=\frac{a^2+c^2-b^2}{2ac}:\frac{2S}{ac}=\frac{a^2+c^2-b^2}{4S}\)

\(\cot C=\frac{cosC}{\sin C}=\frac{a^2+b^2-c^2}{2ab}:\frac{2S}{ab}=\frac{a^2+b^2-c^2}{4S}\)

cot A+cot B+cot C

\(=\frac{b^2+c^2-a^2+a^2+c^2-b^2+a^2+b^2-c^2}{4S}=\frac{a^2+b^2+c^2}{4S}\)

Xét ΔABC có \(cosA=\frac{b^2+c^2-a^2}{2bc};cosB=\frac{a^2+c^2-b^2}{2ac};cosC=\frac{a^2+b^2-c^2}{2ab}\)

Xét ΔABC có \(S_{ABC}=\frac12\cdot bc\cdot\sin A=\frac12\cdot ac\cdot\sin B=\frac12\cdot ab\cdot\sin C\)

=>\(bc\cdot\sin A=ac\cdot\sin B=ab\cdot\sin C=2S\)

=>\(\sin A=\frac{2S}{bc};\sin B=\frac{2S}{ac};\sin C=\frac{2S}{ab}\)

\(\cot A=\frac{cosA}{\sin A}=\frac{b^2+c^2-a^2}{2bc}:\frac{2S}{bc}=\frac{b^2+c^2-a^2}{4S}\)

\(\cot B=\frac{cosB}{\sin B}=\frac{a^2+c^2-b^2}{2ac}:\frac{2S}{ac}=\frac{a^2+c^2-b^2}{4S}\)

\(\cot C=\frac{cosC}{\sin C}=\frac{a^2+b^2-c^2}{2ba}:\frac{2S}{ab}=\frac{a^2+b^2-c^2}{4S}\)

cot A+cot B+cot C

\(=\frac{b^2+c^2-a^2+a^2+c^2-b^2+a^2+b^2-c^2}{4S}\)

\(=\frac{a^2+b^2+c^2}{4S}=\frac{AB^2+AC^2+BC^2}{4S}\)

Sửa đề: \(\cot A+\cot B+\cot C\ge\sqrt3\)

\(\cot A + \cot B = \frac{\cos A}{\sin A} + \frac{\cos B}{\sin B} = \frac{\sin(A + B)}{\sin A \sin B} = \frac{\sin C}{\sin A \sin B}\)

Ta có: \(\cot A \cot B + \cot B \cot C + \cot C \cot A = 1\)

Đặt x=cot A; y=cot B; z=cot C

Ta có: \(\left(x-y\right)^2+\left(y-z\right)^2+\left(x-z\right)^2\ge0\forall x,y,z\)

=>\(x^2+y^2-2xy+y^2+z^2-2yz+x^2+z^2-2xz\ge0\)

=>\(2\left(x^2+y^2+z^2\right)\ge2\left(xy+yz+xz\right)\)

=>\(x^2+y^2+z^2\ge xy+yz+xz\)

=>\(x^2+y^2+z^2+2\left(xy+yz+xz\right)\ge xy+yz+xz+2\left(xy+yz+xz\right)\)

=>\(\left(x+y+z\right)^2\ge3\left(xy+yz+xz\right)\)

=>\((\cot A + \cot B + \cot C)^2 \ge 3(\cot A \cot B + \cot B \cot C + \cot C \cot A) = 3 \cdot 1 = 3\)

=>\(\cot A + \cot B + \cot C \ge \sqrt{3}\)

Dấu '=' xảy ra khi \(\hat{A}=\hat{B}=\hat{C}=\frac{180^0}{3}=60^0\)