Có: \(\left(a+b+c+d\right)^2\ge4\left(a+c\right)\left(b+d\right)=4\left(ab+bc+cd+da\right)=4\)
\(\Rightarrow\left(a+b+c+d\right)^2\ge4\Leftrightarrow a+b+c+d\ge2\)
\(\Rightarrow\dfrac{a^3}{b+c+d}+\dfrac{b+c+d}{8}+\dfrac{a}{6}+\dfrac{1}{12}\ge\dfrac{2a}{3}\left(1\right)\)
Tương tự: \(\dfrac{b^3}{a+c+d}+\dfrac{a+c+d}{8}+\dfrac{b}{6}+\dfrac{1}{12}\ge\dfrac{2b}{3}\left(2\right)\)
\(\dfrac{c^3}{a+b+d}+\dfrac{a+b+d}{8}+\dfrac{c}{6}+\dfrac{1}{12}\ge\dfrac{2c}{3}\left(3\right)\)
\(\dfrac{d^3}{a+b+c}+\dfrac{a+b+c}{8}+\dfrac{d}{6}+\dfrac{1}{12}\ge\dfrac{2d}{3}\left(4\right)\)
Cộng \(\left(1\right)+\left(2\right)+\left(3\right)+\left(4\right)\), có:
\(\Rightarrow A\ge\dfrac{2a}{3}+\dfrac{2b}{3}+\dfrac{2c}{3}+\dfrac{2d}{3}=\dfrac{a+b+c+d}{3}-\dfrac{1}{3}=\dfrac{2}{3}-\dfrac{1}{3}=\dfrac{1}{3}\left(đpcm\right)\)
