Giải thích các bước giải:
$\dfrac{a^{2}}{b}+\dfrac{b^{2}}{c}+\dfrac{4c^{2}}{a}\\
=(\dfrac{a^{2}}{b}+b)+(\dfrac{b^{2}}{c}+4c)+(\dfrac{4c^{2}}{a}+a)-a-b-4c\\
\geq 2\sqrt[]{\dfrac{a^{2}}{b}.b}+2\sqrt[]{\dfrac{b^{2}}{c}.4c}+2\sqrt[]{\dfrac{4c^{2}}{a}.a}-a-b-4c\\
=2a+4b+4c-a-b-4c\\
=a+3b$