Giải thích các bước giải:
Ta có :
$\dfrac{a^3b^3}{c^3(a^2+b^2)}$
$=\dfrac{ab}{c^3(\dfrac{1}{a^2}+\dfrac{1}{b^2})}$
$=\dfrac{abc}{c^4(\dfrac{1}{a^2}+\dfrac{1}{b^2})}$
$=\dfrac{1}{c^4(\dfrac{1}{a^2}+\dfrac{1}{b^2})}$
$=\dfrac{(\dfrac{1}{c^2})^2}{\dfrac{1}{a^2}+\dfrac{1}{b^2}}$
Tương tự
$\dfrac{b^3c^3}{a^3(b^2+c^2)}=\dfrac{(\dfrac{1}{a^2})^2}{\dfrac{1}{b^2}+\dfrac{1}{c^2}}$
$\dfrac{c^3a^3}{b^3(c^2+a^2)}=\dfrac{(\dfrac{1}{b^2})^2}{\dfrac{1}{c^2}+\dfrac{1}{a^2}}$
$\to A=\dfrac{a^3b^3}{c^3(a^2+b^2)}+\dfrac{b^3c^3}{a^3(b^2+c^2)}+\dfrac{c^3a^3}{b^3(c^2+a^2)}$
$\to A=\dfrac{(\dfrac{1}{c^2})^2}{\dfrac{1}{a^2}+\dfrac{1}{b^2}}+\dfrac{(\dfrac{1}{a^2})^2}{\dfrac{1}{b^2}+\dfrac{1}{c^2}}+\dfrac{(\dfrac{1}{b^2})^2}{\dfrac{1}{c^2}+\dfrac{1}{a^2}}$
$\to A\ge \dfrac{(\dfrac{1}{c^2}+\dfrac{1}{a^2}+\dfrac{1}{b^2})^2}{\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+\dfrac{1}{c^2}+\dfrac{1}{a^2}}$
$\to A\ge \dfrac{(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2})^2}{2(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2})}$
$\to A\ge \dfrac{1}{2}(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2})$
$\to A\ge \dfrac{1}{2}.3\sqrt[3]{\dfrac{1}{a^2}.\dfrac{1}{b^2}.\dfrac{1}{c^2}}$
$\to A\ge \dfrac{3}{2}, abc=1$
