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[ongtrum]

cho a,b,c>0,abc=1.CMR: $\dfrac{1}{a^{3}(b+c)}+\dfrac{1}{b^{3}(c+a)}+\dfrac{1}{c^{3}(a+b)}+\dfrac{4(ab+bc+ca)}{(a+b)(b+c)(c+a)} \geq ab+bc+ca$

Edited by ongtrum, 29-05-2008 - 01:07.

[H.Quân- ĐHV]

cho a,b,c>0,abc=1.CMR: $\dfrac{1}{a^{3}(b+c)}+\dfrac{1}{b^{3}(c+a)}+\dfrac{1}{c^{3}(a+b)}+\dfrac{4(ab+bc+ca)}{(a+b)(b+c)(c+a)} \geq ab+bc+ca$

Cũng không khó lắm
đặt $\dfrac{1}{a} = x , \dfrac{1}{b} = y , \dfrac{1}{c} = z $thế thì
$\dfrac{1}{a^{3}(b+c)}+\dfrac{1}{b^{3}(c+a)}+\dfrac{1}{c^{3}(a+b)}+\dfrac{4(ab+bc+ca)}{(a+b)(b+c)(c+a)}$ = $ \sum \dfrac{x^2}{y+z} + \dfrac{4(x+y+z)}{(x+y)(y+z)(x+z)}$
cần cm $ \sum \dfrac{x^2}{y+z} + \dfrac{4(x+y+z)}{(x+y)(y+z)(x+z)} \ge x+y+z$
ta có $ \dfrac{x^2}{y+z} + x = \dfrac{x(x+y+z)}{y+z}$
vậy chỉ cần cm$ \sum \dfrac{x}{y+z} + \dfrac{4}{(x+y)(y+z)(x+z)} \ge 2$ $x^3+y^3+z^3 +3xyz \ge \sum x^2y$ (đúng)

Edited by H.Quân- ĐHV, 06-06-2008 - 10:45.