Trích:
Nguyên văn bởi kaichana  10. Cho x, y, z >0. CMR:
$$\frac{{{x^2}z}}{{xyz + {y^3}}} + \frac{{{y^2}x}}{{xyz + {z^3}}} + \frac{{{z^2}y}}{{xyz + {x^3}}} \ge \frac{1}{2}\left( {\frac{x}{y} + \frac{y}{z} + \frac{z}{x}} \right)$ $ |
Ta đặt: $\frac{x}{y}=a;\frac{y}{z}=b;\frac{z}{x}=c $
Khi đó,
VT = $\frac{{{x^2}z}}{{xyz + {y^3}}} + \frac{{{y^2}x}}{{xyz + {z^3}}} + \frac{{{z^2}y}}{{xyz + {x^3}}}
=\frac{1}{\frac{y}{x}+\frac{y}{z}.\frac{y^{2}}{x^{ 2}}}+\frac{1}{\frac{z}{y}+\frac{z}{x}.\frac{z^{2}} {y^{2}}}+\frac{1}{\frac{x}{z}+\frac{x}{y}.\frac{x^ {2}}{z^{2}}}
=\frac{1}{\frac{1}{a}+\frac{b}{a^{2}}}+\frac{1}{\f rac{1}{b}+\frac{c}{b^{2}}}+\frac{1}{\frac{1}{c}+\f rac{a}{c^{2}}}
=\frac{a^{2}}{a+b}+\frac{b^{2}}{b+c}+\frac{c^{2}}{ c+a}
\geq \frac{1}{2}.(a+b+c) $
Đúng theo BĐT Cauchy-Schward. Đẳng thức xảy ra khi và chỉ khi
a = b = c ,hay: x = y =z. $\square $
[RIGHT][I][B]Nguồn: MathScope.ORG[/B][/I][/RIGHT]