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$\sum a\sqrt{\frac{a}{b(b+3)}}\geq \frac{3}{2}$


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#1
beontop97

beontop97

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Cho a,b,c>0 thỏa a+b+c=3

Chứng minh:

$a\sqrt{\frac{a}{b(b+3)}}+b\sqrt{\frac{b}{c(c+3)}}+c\sqrt{\frac{c}{a(a+3)}}\geq \frac{3}{2}$

 


Edited by beontop97, 17-10-2014 - 14:11.


#2
dogsteven

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$VT=\sum \dfrac{2a^2}{\sqrt{4ab(b+3)}} \geqslant \sum \dfrac{4a^2}{4ab+b+3} \geqslant \dfrac{9}{ab+bc+ca+3} \geqslant \dfrac{3}{2}$


Quyết tâm off dài dài cày hình, số, tổ, rời rạc.





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