Xét \(3a^4-4a^3+1=\left(a-1\right)^2\left(3a^2+2a+1\right)\ge0\)
\(2b^3-3b^2+1=\left(2b+1\right)\left(b-1\right)^2\ge0;\forall b>0\)
\(c^2-2c+1=\left(c-1\right)^2\ge0\)
\(\Rightarrow\left(3a^4+2b^3+c^2\right)-\left(4a^3+3b^2+2c\right)+3\ge0\)
\(\Rightarrow3a^4+2b^3+c^2\ge4a^3+3b^2+2c-3=6\)
Dấu "=" xảy ra khi a=b=c=1
