Ta có:
$1 + a^2 = ab + bc + ca + a^2$
$= b(a +c) + a(a + c) = (a + b)(b + c)$
$\Rightarrow \sqrt{1 + a^2} = \sqrt{(a + b)(b + c)}$
Áp dụng bất đẳng thức $AM-GM$ ta được:
$\dfrac{\dfrac{1}{a + b} + \dfrac{1}{b+ c}}{2} \geq \dfrac{1}{\sqrt{a + b}}\cdot\dfrac{1}{\sqrt{b + c}}$
$\Leftrightarrow \dfrac{\dfrac{a}{a + b} + \dfrac{a}{b+ c}}{2} \geq \dfrac{a}{\sqrt{(a + b)(b + c)}}$
Tương tự, ta được:
$\dfrac{\dfrac{b}{b + c} + \dfrac{b}{b+ a}}{2} \geq \dfrac{b}{\sqrt{(b+c)(b + a)}}$
$\dfrac{\dfrac{c}{c + a} + \dfrac{c}{c+ b}}{2} \geq \dfrac{c}{\sqrt{(c+b)(c + a)}}$
Cộng vế theo vế, ta được:
$\dfrac{\dfrac{a}{a + b} + \dfrac{a}{b+ c} + \dfrac{b}{b + c} + \dfrac{b}{b+ a} +\dfrac{c}{c + a} + \dfrac{c}{c+ b} }{2} \geq \dfrac{a}{\sqrt{(a + b)(b + c)}} + \dfrac{b}{\sqrt{(b+c)(b + a)}} + \dfrac{c}{\sqrt{(c+b)(c + a)}}$
$\Leftrightarrow \dfrac{3}{2} \geq \dfrac{a}{1 + a^2} + \dfrac{b}{1 + b^2} + \dfrac{c}{1 + c^2}$
Dấu = xảy ra $\Leftrightarrow a = b = c = \dfrac{\sqrt3}{3}$
