Ta có: $2x^2 - 11x + 21 > 0 \Rightarrow 3\sqrt[3]{4x-4} >0 \Rightarrow x-1 >0$Theo $\mbox{AM-GM}$, ta có: $2\left(x-1\right)^2 +8 \ge 8\left(x-1\right)$
Tương tự: $\left(x-1\right)+2+2 \ge 3\sqrt[3]{2\times2\times\left(x-1\right)} = 3\sqrt[3]{4x-4}$
Suy ra: $2x^2 -11x + 21 - 3\sqrt[3]{4x-4} \ge 0$Dấu $"="$ xảy ra $\Leftrightarrow x-1 = 2 \Rightarrow x = 3.\,\,\,\blacksquare$