Question

Cho x,y,z dương thảo mãn: \(xy+yz+zx=671\) . CMR

\(\frac{x}{x^2-yz+2013}+\frac{y}{y^2-xz+2013}+\frac{z}{z^2-xy+2013}\ge\frac{1}{x+y+z}\)

Answer 1

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Answer 2

\(\frac{x^2}{x^3-xyz+2013x}+\frac{y^2}{y^3-xyz+2013y}+\frac{z^2}{z^3-xyz+2013z}\)

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

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

Answer 3

\(VT=\text{Σ}_{cyc}\frac{x}{x^2-yz+2013}=\text{Σ}_{cyc}\frac{x^2}{x^3-xyz+2013x}\)

\(\ge\frac{\left(x+y+z\right)^2}{x^3+y^3+z^3-3xyz+2013\left(x+y+z\right)}\)(bđt Cauchy - Schwarz dạng Engel)

\(=\frac{\left(x+y+z\right)^2}{\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)+2013\left(x+y+z\right)}\)

\(=\frac{\left(x+y+z\right)^2}{\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx+2013\right)}\)

\(=\frac{\left(x+y+z\right)^2}{\left(x+y+z\right)\left[\left(x+y+z\right)^2-3\left(xy+yz+zx\right)+2013\right]}\)

\(=\frac{\left(x+y+z\right)^2}{\left(x+y+z\right)\left[\left(x+y+z\right)^2-3.671+2013\right]}\)

\(=\frac{\left(x+y+z\right)^2}{\left(x+y+z\right)^3}=\frac{1}{x+y+z}\)

(Dấu "=" xảy ra khi x = y = z = \(\frac{\sqrt{2013}}{3}\))