$VT=abc\sum_{}^{}\frac{1}{a^4(2b+c)}=\frac{a+b+c}{3}.\sum_{}^{} \frac{\frac{1}{a^4}}{2b+c}$$\geq \frac{a+b+c}{3}.\frac{(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2})^2}{3(a+b+c)}$
$\geq \frac{(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca})^2}{9}=\frac{3^2}{9}=1.$
Dấu $=$ xảy ra $\Leftrightarrow a=b=c=1$.
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