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[onlyloveyouonly]

cho a,b,c>0.CMR: $ \sqrt{ \dfrac{b+c}{a}} +\ sqrt{ \dfrac{c+a}{b}}+\ sqrt{\dfrac{a+b}{c }}\geq sqrt{\dfrac{6(a+b+c)}{\sqrt[3]{abc}}} $

Edited by onlyloveyouonly, 25-05-2008 - 12:30.

[quangghePT1]

cho a,b,c>0.CMR: $ \sqrt{ \dfrac{b+c}{a}} +\ sqrt{ \dfrac{c+a}{b}}+\ sqrt{\dfrac{a+b}{c }}\geq sqrt{\dfrac{6(a+b+c)}{\sqrt[3]{abc}}} $

Chuẩn hóa $a+b+c=1$

Cần c/m

$\sum \sqrt{bc(b+c)} \geq \sqrt{6} \sqrt[3]{abc}$

$VT=\sqrt{b^2c+c^2b}+\sqrt{a^2b+b^2a}+\sqrt{c^2a+a^2c}\geq \sqrt{(b\sqrt{c}+a\sqrt{b}+c\sqrt{a})^2+(c\sqrt{b}+b\sqrt{a}+a\sqrt{c})^2}\geq 3\sqrt{2}\sqrt[4]{abc}=3\sqrt{2}\dfrac{\sqrt[3]{abc}}{\sqrt[12]{abc}}\geq \sqrt{6}\sqrt[3]{abc}$

Edited by quangghePT1, 26-05-2008 - 21:22.