Lấy điểm O trong ΔABC sao cho \(\hat{AOB}=\hat{BOC}=\hat{AOC}=120^0\)

Đặt OA=x; OB=y; OC=z

Xét ΔOAB có \(cosAOB=\frac{OA^2+OB^2-AB^2}{2\cdot OA\cdot OB}\)

=>\(x^2+y^2-AB^2=2\cdot x\cdot y\cdot\frac{-1}{2}=-xy\)

=>\(AB^2=x^2+y^2+xy=1\)

=>AB=1

Xét ΔOAC có \(cosAOC=\frac{OA^2+OC^2-AC^2}{2\cdot OA\cdot OC}\)

=>\(x^2+z^2-AC^2=2\cdot x\cdot z\cdot cos120=-xz\)

=>\(AC^2=x^2+z^2+xz=\frac34\)

=>\(AC=\frac{\sqrt3}{2}\)

Xét ΔOBC có \(cosBOC=\frac{OB^2+OC^2-BC^2}{2\cdot OB\cdot OC}\)

=>\(\frac{y^2+z^2-BC^2}{2\cdot y\cdot z}=cos120=-\frac12\)

=>\(y^2+z^2-BC^2=-yz\)

=>\(BC^2=y^2+z^2+yz=\frac14\)

=> BC=1/2

Vì \(CA^2+CB^2=AB^2\)

nên ΔCAB vuông tại C

=>\(S_{CAB}=\frac12\cdot CA\cdot CB=\frac12\cdot\frac12\cdot\frac{\sqrt3}{2}=\frac{\sqrt3}{8}\)

\(S_{ABC} = \frac{1}{2}xy\sin(120^\circ) + \frac{1}{2}yz\sin(120^\circ) + \frac{1}{2}xz\sin(120^\circ)\)

=>\(S_{ABC} = \frac{\sqrt{3}}{4}(xy + yz + xz)\)

=>\(\frac{\sqrt{3}}{4}(xy+yz+xz)=\frac{\sqrt{3}}{8}\)

=>\(xy+yz+xz=\frac{1}{2}\)

\((x^2 + xy + y^2) + (y^2 + yz + z^2) + (x^2 + xz + z^2) = 1 + \frac{1}{4} + \frac{3}{4}\)

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

=>\(2(x^2+y^2+z^2)+\frac{1}{2}=2\)

=>\(x^2+y^2+z^2=\frac{3}{4}\)

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

\(=\frac34+2\cdot\frac12=\frac74\)

=>\(x+y+z=\frac{\sqrt7}{2}\)

=>\(B=\frac{\sqrt7}{2}\)

thêm x²+y²+z²=1 nha

thêm x² + y² + z² = 1 nha

      HT nha vinh

\(xy+yz+zx=4xyz\Rightarrow\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=4\)

Ta có \(M=\frac{1}{4\left(x+y\right)}+\frac{1}{4\left(y+z\right)}+\frac{1}{4\left(z+x\right)}\)

               \(=\frac{1}{16}\left(\frac{4}{x+y}+\frac{4}{y+z}+\frac{4}{z+x}\right)\le\frac{1}{16}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{y}+\frac{1}{z}+\frac{1}{z}+\frac{1}{x}\right)\)

                                                                                     \(=\frac{1}{8}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)=\frac{1}{8}.4=\frac{1}{2}\)

Dấu "=" tại x = y = z = ³/₄