Ta có:

$\dfrac{1}{ab} + \dfrac{1}{bc} + \dfrac{1}{ca}$

$= \dfrac{ab+ bc+ ca}{ab} + \dfrac{ab+ bc+ ca}{bc} + \dfrac{ab+ bc+ ca}{ca}$

$= 1 + \dfrac{bc + ca}{ab} + 1 + \dfrac{ab + ca}{bc} + 1 + \dfrac{ab + bc}{ca}$

$= 3 + \dfrac{c}{a} + \dfrac{c}{b} + \dfrac{a}{c} + \dfrac{a}{b} + \dfrac{b}{c} + \dfrac{b}{a}$

Mặt khác:

$\sqrt{\dfrac{(a + b)(a + c)}{a^2}} \leq \dfrac{a +b + a + c}{2a} = \dfrac{2a + b + c}{2a} = 1 + \dfrac{1}{2}\left(\dfrac{b}{a} + \dfrac{c}{a}\right)$

Tương tự:

$\sqrt{\dfrac{(b + c)(b + a)}{b^2}} \leq 1 + \dfrac{1}{2}\left(\dfrac{c}{b} + \dfrac{a}{b}\right)$

$\sqrt{\dfrac{(c + a)(c+ b)}{c^2}} \leq 1 + \dfrac{1}{2}\left(\dfrac{a}{c} + \dfrac{b}{c}\right)$

$\Rightarrow 3 + \sqrt{\dfrac{(a + b)(a + c)}{a^2}} + \sqrt{\dfrac{(b + c)(b + a)}{b^2}} \sqrt{\dfrac{(c + a)(c+ b)}{c^2}} \leq 6 + \dfrac{1}{2}\left(\dfrac{c}{a} + \dfrac{c}{b} + \dfrac{a}{c} + \dfrac{a}{b} + \dfrac{b}{c} + \dfrac{b}{a}\right)$

Bất đẳng thức đã cho trở thành:

$3 + \dfrac{c}{a} + \dfrac{c}{b} + \dfrac{a}{c} + \dfrac{a}{b} + \dfrac{b}{c} + \dfrac{b}{a} \geq 6 + \dfrac{1}{2}\left(\dfrac{c}{a} + \dfrac{c}{b} + \dfrac{a}{c} + \dfrac{a}{b} + \dfrac{b}{c} + \dfrac{b}{a}\right)$

$\Leftrightarrow \dfrac{1}{2}\left(\dfrac{c}{a} + \dfrac{c}{b} + \dfrac{a}{c} + \dfrac{a}{b} + \dfrac{b}{c} + \dfrac{b}{a}\right) \geq 3$ $(*)$

Áp dụng bất đẳng thức $AM-GM$ ta được:

$\dfrac{1}{2}\left(\dfrac{c}{a} + \dfrac{c}{b} + \dfrac{a}{c} + \dfrac{a}{b} + \dfrac{b}{c} + \dfrac{b}{a}\right) \geq \dfrac{1}{2}.6\sqrt[6]{\dfrac{c}{a}\cdot\dfrac{c}{b}\cdot\dfrac{a}{c}\cdot\dfrac{a}{b}\cdot\dfrac{b}{c}\cdot\dfrac{b}{a}} = \dfrac{1}{2}.6 = 3$

$\Rightarrow (*)$ đúng

Vậy bất đẳng thức được chứng minh