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[eminemdech]

Cho các số thực dương a,b,c thõa mãn $abc\doteq \frac{1}{6}$.

Chứng minh $3+\frac{a}{2b}+\frac{2b}{3c}+\frac{3c}{a}\geq a+2b+3c+\frac{1}{a}+\frac{1}{2b}+\frac{1}{3c}$

 

Edited by hachinh2013, 18-02-2015 - 09:43.

[Dinh Xuan Hung]

$\left\{\begin{matrix} a=\frac{y}{x} & & & \\ 2b=\frac{z}{y}& & & \\ 3c=\frac{x}{z}& & & \end{matrix}\right.$

$3+\frac{a}{2b}+\frac{2b}{3c}+\frac{3c}{a}\geq a+2b+3c+\frac{1}{a}+\frac{1}{2b}+\frac{1}{3c}\Leftrightarrow 3+\sum \frac{y^2}{xz}\geq \sum \frac{y}{x}+\sum \frac{x}{y}\Leftrightarrow x^3+y^3+z^3+3xyz\geq y^2z+y^2x+x^2z+x^2y+z^2x+z^2y\Leftrightarrow \sum x(x-y)(x-z)\geq 0$

(Luôn đúng theo BĐT SHUR )

Edited by Dinh Xuan Hung, 18-02-2015 - 15:29.