$P = \dfrac{ab}{c^2(a+ b)} + \dfrac{bc}{a^2(b+c)} + \dfrac{ca}{b^2(c + a)}$
$\Leftrightarrow P = \dfrac{a^2b^2}{abc(ca + cb)} + \dfrac{b^2c^2}{abc(ab + ac)} + \dfrac{c^2a^2}{abc(bc + ba)}$
Áp dụng bất đẳng thức $Cauchy-Schwarz$ dạng $Engel$ ta được:
$\dfrac{a^2b^2}{abc(ca + cb)} + \dfrac{b^2c^2}{abc(ab + ac)} + \dfrac{c^2a^2}{abc(bc + ba)} \geq \dfrac{(ab +bc + ca)^2}{2abc(ab + bc + ca)} = \dfrac{ab + bc + ca}{2abc}$
$\geq \dfrac{3\sqrt[3]{a^2b^2c^2}}{2abc} = \dfrac{3}{2\sqrt[3]{abc}}$
$\geq \dfrac{3}{2.\dfrac{a+b+c}{3}} = \dfrac{3}{2.\dfrac{3}{3}} = \dfrac{3}{2}$
Dấu = xảy ra $\Leftrightarrow a = b = c = 1$
