Bài 3:ÁD BĐT Cauchy-Schwartz:
$ \frac{a}{2a^{2}+bc}+\frac{b}{2b^{2}+ca}+\frac{c}{2c^{2}+ab}=\frac{1}{2a+\frac{bc}{a}}+\frac{1}{2b+\frac{ca}{b}}+\frac{1}{2c+\frac{ab}{c}}$
$\geq \frac{9}{2(a+b+c)+(\frac{bc}{a}+\frac{ca}{b}+\frac{ab}{c})}=\frac{9}{2(a+b+c)+\frac{a^{2}b^{2}+b^{2}c^{2}+c^{2}a^{2}}{abc}}$
=$\frac{9abc}{2abc(a+b+c)+(a^{2}b^{2}+b^{2}c^{2}+c^{2}a^{2}}=\frac{9abc}{(ab+bc+ca)^{2}}=abc( ab+bc+ca=3)$
$\Rightarrow$đpcm
Dấu''='' xra$\Leftrightarrow a=b=c=1$