Đặt $P = \dfrac{1}{a^3.(b+c)} + \dfrac{1}{b^3.(a+c)} + \dfrac{1}{c^3.(a+b)}$
$ = \dfrac{\dfrac{1}{a^2}}{a.(b+c)} + \dfrac{\dfrac{1}{b^2}}{b.(a+c)} + \dfrac{\dfrac{1}{c^2}}{c.(a+b)}$
Áp dụng BĐT Svac - xơ ta có :
$ P= \dfrac{\dfrac{1}{a^2}}{a.(b+c)} + \dfrac{\dfrac{1}{b^2}}{b.(a+c)} + \dfrac{\dfrac{1}{c^2}}{c.(a+b)}$
$≥ \dfrac{\bigg(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\bigg)^2}{2.(ab+bc+ca)}$
$ = \dfrac{\dfrac{(ab+bc+ca)^2}{abc}}{2.(ab+bc+ca)}$
$ = \dfrac{ab+bc+ca}{2} ≥ \dfrac{3\sqrt[3]{ab.bc.ca}}{2} = \dfrac{3}{2}$
Dấu "=" xảy ra $⇔a=b=c=1$
Vậy BĐT được chứng minh.
