Áp dụng Bất đẳng thức $Cauchy$ ta có: +) $\frac{a^2}{(a-1)^2}+\frac{64}{27}.\frac{a-1}{a}+\frac{64}{27}.\frac{a-1}{a}\geq \frac{16}{3}$+) $\frac{b^2}{(b-1)^2}+\frac{1}{27}.\frac{b-1}{b}+\frac{1}{27}.\frac{b-1}{b}\geq \frac{1}{3}$+) $\frac{c^2}{(c-1)^2}+\frac{1}{27}.\frac{c-1}{c}+\frac{1}{27}.\frac{c-1}{c}\geq \frac{1}{3}$Cộng lại theo vế ta có: $VT+\frac{128}{27}.(1-\frac{1}{a})+\frac{2}{27}.(1-\frac{1}{b})+\frac{2}{27}.(1-\frac{1}{c})\geq 6$$VT+\frac{44}{9}-\frac{2}{27}.(\frac{64}{a}+\frac{1}{b}+\frac{1}{c})\geq 6$$VT\geq \frac{10}{9}+\frac{2}{27}.(\frac{(-8)^2}{a}+\frac{1}{b}+\frac{1}{c})$Áp dụng bất đẳng thức $Cauchy-Schwart$ ta có: $VT\geq \frac{10}{9}+\frac{2}{27}.\frac{(-8+1+1)^2}{a+b+c}=2$Dấu bằng xảy ra khi $a=4 ; b=c=\frac{-1}{2}$

Áp dụng Bất đẳng thức $Cauchy$ ta có: +) $\frac{a^2}{(a-1)^2}+\frac{64}{27}.\frac{a-1}{a}+\frac{64}{27}.\frac{a-1}{a}\geq \frac{16}{3}$+) $\frac{b^2}{(b-1)^2}+\frac{1}{27}.\frac{b-1}{b}+\frac{1}{27}.\frac{b-1}{b}\geq \frac{1}{3}$+) $\frac{c^2}{(c-1)^2}+\frac{1}{27}.\frac{c-1}{c}+\frac{1}{27}.\frac{c-1}{c}\geq \frac{1}{3}$Cộng lại theo vế ta có: $VT+\frac{128}{27}.(1-\frac{1}{a})+\frac{2}{27}.(1-\frac{1}{b})+\frac{2}{27}.(1-\frac{1}{c})\geq 6$$VT+\frac{44}{9}-\frac{2}{27}.(\frac{64}{a}+\frac{1}{b}+\frac{1}{c})\geq 6$$VT\geq \frac{10}{9}+\frac{2}{27}.(\frac{(-8)^2}{a}+\frac{1}{b}+\frac{1}{c})$Áp dụng bất đẳng thức $Cauchy-Schwart$ ta có: $VT\geq \frac{10}{9}+\frac{2}{27}.\frac{(-8+1+1)^2}{a+b+c}=2$Dấu bằng xảy ra khi $a=4 ; b=c=\frac{-1}{2}$