$\begin{array}{l}\text{Bạn bổ sung điều kiện $a,b,c>0$}\\(a^3+b^3+c^3)\left(\dfrac1a+\dfrac1b+\dfrac1c\right)=a^2+\dfrac{a^3}b+\dfrac{a^3}c+\dfrac{b^3}a+b^2+\dfrac{b^3}c+\dfrac{c^3}a+\dfrac{c^3}b+c^2\\=a^2+b^2+c^2+\left(\dfrac{a^3}b+\dfrac{b^3}a\right)+\left(\dfrac{b^3}c+\dfrac{c^3}b\right)+\left(\dfrac{c^3}a+\dfrac{a^3}c\right)\\\text{Áp dụng BĐT Cô-si cho các số dương, ta có:}\\\left(\dfrac{a^3}b+\dfrac{b^3}a\right)+\left(\dfrac{b^3}c+\dfrac{c^3}b\right)+\left(\dfrac{c^3}a+\dfrac{a^3}c\right)\ge 2ab+2bc+2ca\\\to a^2+b^2+c^2+\left(\dfrac{a^3}b+\dfrac{b^3}a\right)+\left(\dfrac{b^3}c+\dfrac{c^3}b\right)+\left(\dfrac{c^3}a+\dfrac{a^3}c\right)\ge a^2+b^2+c^2+2ab+2bc+2ca=(a+b+c)^2\\\text{Đẳng thức xảy ra} \ \leftrightarrow a=b=c\\\text{Vậy BĐT được chứng minh.}\end{array}$
