Question

Cho x,y,z dương thỏa mãn x + y + z = xy + yz + zx. Chứng minh:

\(\frac{1}{x^2+y+1}+\frac{1}{y^2+z+1}+\frac{1}{z^2+x+1}\le1\)

Answer 1

Đã tìm ra lời giải:

gt \(\Rightarrow\left(xy+yz+zx\right)^2=\left(x+y+z\right)^2\ge3\left(xy+yz+zx\right)\)

\(\Leftrightarrow xy+yz+zx\ge3\)

Áp dụng bđt Bunhiacopxki:

\(\frac{1}{\left(x^2+y+1\right)\left(1+y+z^2\right)}\le\frac{1}{\left(x+y+z\right)^2}\Rightarrow\frac{1}{x^2+y+1}\le\frac{1+y+z^2}{\left(x+y+z\right)^2}\)

Tương tự rồi cộng lại, ta được:

\(VT\le\frac{\left(x^2+y^2+z^2\right)+\left(x+y+z\right)+3}{\left(x+y+z\right)^2}\)

\(=\frac{\left(x+y+z\right)^2-2\left(xy+yz+zx\right)+\left(xy+yz+zx\right)+3}{\left(x+y+z\right)^2}\)

\(=1+\frac{-\left(xy+yz+zx\right)+3}{\left(xy+yz+zx\right)^2}\le1+\frac{-3+3}{3^2}=1\)

Dấu đẳng thức xảy ra khi x = y = z = 1