Question

tìm min P=1/(1+xy)+1/(1+yz)+1/(1+xz), trong đó x, y, z thỏa mãn x^2+y^2+z^2 < hoặc = 3

Answer 1

\(xy+yz+xz\le x^2+y^2+z^2\le3\)

\(\frac{1}{1+xy}+\frac{1}{1+yz}+\frac{1}{1+xz}\ge\frac{9}{3+xy+yz+zx}\ge\frac{9}{3+3}=\frac{3}{2}\)

Dấu"=" xảy ra khi x=y=z=1

Vậy...

Answer 2

MIn=3/2 khi x=y=z=1

Answer 3

\(P\ge\frac{9}{1+xy+1+yz+1+zx}=\frac{9}{3+\left(xy+yz+zx\right)}\)

Mà \(xy+yz+zx\le x^2+y^2+z^2\le3\)

\(P\ge\frac{9}{3+3}=\frac{3}{2}\)

Dấu bằng xảy ra khi x=y=z=1