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[Xuan Hung HQH]

Cho a,b,c dương TM a+b+c=6abc.CMR:$\frac{bc}{a^3(c+2b)}+\frac{ac}{b^3(a+2c)}+\frac{ab}{c^3(b+2a)}\geq 2$

[Nguyen Duc Thuan]

Cho a,b,c dương TM a+b+c=6abc.CMR:$\frac{bc}{a^3(c+2b)}+\frac{ac}{b^3(a+2c)}+\frac{ab}{c^3(b+2a)}\geq 2$

Từ GT ta có: $\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}=6$

Đặt $\sum \frac{1}{a}=\sum x\Rightarrow xy+yz+zx=6$

Áp dụng  BĐT Cauchy-Schawrz ta có:

$P=\sum \frac{x^3}{y+2z}=\sum \frac{x^4}{xy+2zx}\geq \frac{(x^2+y^2+x^2)^2}{3(xy+yz+zx)}\geq \frac{x^2+y^2+z^2}{3}\geq 2$

(do $x^2+y^2+z^2\geq xy+yz+zx=6$)

Dấu "=" xảy ra khi a=b=c=1/2

Edited by Nguyen Duc Thuan, 15-11-2014 - 01:39.