1 reply to this topic

[Leuleudoraemon]

a,b,c>0 tm $abc=1$

 

CM:

       $\sum \frac{b^2}{a}+\frac{9}{2(ab+bc+ca)}\geq \frac{9}{2}$

Edited by Leuleudoraemon, 01-04-2018 - 23:13.

[PugMath]

a,b,c>0 tm $abc=1$

 

CM:

       $\sum \frac{b^2}{a}+\frac{9}{2(ab+bc+ca)}\geq \frac{9}{2}$

$VT\geqslant a+b+c+\frac{9}{2(ab+bc+ac)}\geqslant \sqrt{3(ab+bc+ac)}+\frac{9}{2(ab+bc+ac)}=\frac{\sqrt{3(ab+bc+ac)}}{2}+\frac{\sqrt{3(ab+bc+ac)}}{2}+\frac{9}{2(ab+bc+ac)}\geqslant \frac{9}{2}$

điều kiện thừa quá