Ta có
$\sqrt{\dfrac{ab}{c+ab}} = \sqrt{\dfrac{ab}{c.1+ab}}$
$= \sqrt{\dfrac{ab}{c(a+b+c) + ab}}$
$= \sqrt{\dfrac{ab}{c(a+c) + bc + ab}}$
$= \sqrt{\dfrac{ab}{c(a+c) + b(a+c)}}$
$= \sqrt{\dfrac{ab}{(a+c)(b+c)}}$
$= \sqrt{\dfrac{a}{a+c} . \dfrac{b}{b+c}} \leq \dfrac{1}{2} \left( \dfrac{a}{a+c} + \dfrac{b}{b+c}\right)$
Tương tự ta cx có
$\sqrt{\dfrac{bc}{a+bc}} \leq \dfrac{1}{2} \left( \dfrac{b}{a+b} + \dfrac{c}{c+a} \right)$
$\sqrt{\dfrac{ca}{b+ca}} \leq \dfrac{1}{2} \left( \dfrac{c}{b+c} + \dfrac{a}{a+b} \right)$
Khi đó, ta có
$P \leq \dfrac{1}{2} \left( \dfrac{a}{a+c} + \dfrac{b}{b+c} + \dfrac{b}{a+b} + \dfrac{c}{c+a} + \dfrac{c}{b+c} + \dfrac{a}{a+b} \right)$
Lại có
$\dfrac{a}{a+c} + \dfrac{b}{b+c} + \dfrac{b}{a+b} + \dfrac{c}{c+a} + \dfrac{c}{b+c} + \dfrac{a}{a+b} = \dfrac{a}{a+c} + \dfrac{c}{a+c} + \dfrac{b}{a+b} + \dfrac{a}{a+b} + \dfrac{b}{b+c} + \dfrac{c}{b+c}$
$= 1 + 1 + 1 = 3$
Vậy ta có
$P \leq \dfrac{3}{2}$
Dấu "=" xảy ra khi và chỉ khi $\sqrt{\dfrac{ab}{c+ab}} = \sqrt{\dfrac{bc}{a+bc}} = \sqrt{\dfrac{ca}{b+ca}}$ hay $a = b = c = \dfrac{1}{3}$.
