Áp dụng BĐT Cauchy ta có :
$\dfrac{a^3}{(a+b).(a+c)} + \dfrac{a+b}{8}+\dfrac{a+c}{8} ≥ 3.\sqrt[3]{\dfrac{a^3}{(a+b).(a+c)} .\dfrac{a+b}{8}.\dfrac{a+c}{8}} = \dfrac{3a}{4}$
Tương tự có :
$\dfrac{b^3}{(b+a).(b+c)} + \dfrac{b+a}{8}+\dfrac{b+c}{8} ≥ 3.\sqrt[3]{\dfrac{b^3}{(b+a).(b+c)} .\dfrac{b+a}{8}.\dfrac{b+c}{8}} = \dfrac{3b}{4}$
$\dfrac{c^3}{(c+a).(c+b)} + \dfrac{c+a}{8}+\dfrac{c+b}{8} ≥ 3.\sqrt[3]{\dfrac{c^3}{(c+a).(c+b)} .\dfrac{c+a}{8}.\dfrac{c+b}{8}} = \dfrac{3c}{4}$
Cộng vế với vế của các BĐT trên có :
$\dfrac{a^3}{(a+b).(a+c)} + \dfrac{b^3}{(b+c).(b+a)} + \dfrac{c^3}{(c+a).(c+b)} + \dfrac{4.(a+b+c)}{8} ≥ \dfrac{3}{4}.(a+b+c)$
$\to \dfrac{a^3}{(a+b).(a+c)} + \dfrac{b^3}{(b+c).(b+a)} + \dfrac{c^3}{(c+a).(c+b)} ≥ \dfrac{a+b+c}{4}$
$\to đpcm$
