` P = x/(y+z) +y/(z+x) + z/(x+y) = (x^2)/(xy+xz) + (y^2)/(yz+xy) + (z^2)/(xz + yz)`
Áp dụng BĐT Cauchy - Schwarz dạng Engel ta có
` P \ge ((x+y+z)^2)/(xy + xz + yz + xy + xz + yz) = (( x+y+z)^2)/(2(xy+yz+xz))`
Mặt khác ta có BĐT phụ
` (x+y+z)^2 \ge 3(xy + yz+ xz)`
`\to x^2 +y^2 +z^2 + 2xy + 2yz + 2xz - 3xy - 3yz -3xz \ge 0`
`\to x^2 + y^2 +z^2 - xy-yz - xz \ge 0`
`\to 2x^2 + 2y^2 + 2z^2 - 2xy - 2yz - 2xz \ge 0`
`\to (x-y)^2+(y-z)^2+(x-z)^2 \ge 0`
`\to P \ge (3(xy+yz+xz))/(2(xy+yz+xz)) = 3/2`
Vậy GTNN ` P = 3/2` khi ` x = y = z `