$VT=(a+b+c)\sum\frac{1}{3a+3b+2c}+2\sum \frac{c}{3a+3b+2c}$
$\geq (a+b+c)\frac{9}{8(a+b+c)}+2\sum \frac{a^{2}}{2a^{2}+3ab+3ac}\geq \frac{9}{8}+\frac{2(a+b+c)^{2}}{2\sum a^{2}+6\sum ab}\geq \frac{9}{8}+\frac{2(a+b+c)^{2}}{2(a+b+c)^{2}+\frac{2}{3}(a+b+c)^{2})}=\frac{9}{8}+\frac{3}{4}=\frac{15}{8}$
Giải bằng cauchy-schwarz có lẽ dễ dàng hơn đặt ẩn