Giải thích các bước giải:
Bài 12:
Ta có :
$a(\dfrac{a}{2}+\dfrac{1}{bc})+b(\dfrac{b}{2}+\dfrac{1}{ca})+c(\dfrac{c}{2}+\dfrac{1}{ab})$
$=\dfrac{a^2+b^2+c^2}{2}+\dfrac{a^2+b^2+c^2}{abc}$
$\ge \dfrac{3\sqrt[3]{a^2b^2c^2}}{2}+\dfrac{3\sqrt[3]{a^2b^2c^2}}{abc}$
$\ge 3(\dfrac{\sqrt[3]{a^2b^2c^2}}{2}+\dfrac{\sqrt[3]{a^2b^2c^2}}{abc})$
$\ge 3(\dfrac{\sqrt[3]{a^2b^2c^2}}{2}+\dfrac{\sqrt[3]{a^2b^2c^2}}{2abc}+\dfrac{\sqrt[3]{a^2b^2c^2}}{2abc})$
$\ge 3.3\sqrt[3]{\dfrac{\sqrt[3]{a^2b^2c^2}}{2}.\dfrac{\sqrt[3]{a^2b^2c^2}}{2abc}.\dfrac{\sqrt[3]{a^2b^2c^2}}{2abc}}$
$=\dfrac{9}{2}$
