Áp dụng bất đẳng thức $Cauchy - Schwarz$ dạng $Engel$ ta được:
$\dfrac{1}{a + b} + \dfrac{1}{a + c} \geq \dfrac{(1 +1)^2}{a + b + a + c} = \dfrac{4}{2a + b + c}$
$\Leftrightarrow \dfrac{a}{a + b} + \dfrac{a}{a + c} \geq \dfrac{4a}{2a + b + c}$
$\Leftrightarrow \dfrac{a}{4(a + b)} + \dfrac{a}{4(a + c)} \geq \dfrac{a}{2a + b + c}$
Tương tự ta được:
$\dfrac{b}{4(b + a)} +\dfrac{b}{4(b + c)} \geq \dfrac{b}{2b + c +a}$
$\dfrac{c}{4(c + a)} + \dfrac{c}{4(c + b)} \geq \dfrac{c}{2c + a + b}$
Cộng vế theo vế ta được:
$\dfrac{a}{4(a + b)} + \dfrac{a}{4(a + c)} + \dfrac{b}{4(b + a)} +\dfrac{b}{4(b + c)} + \dfrac{c}{4(c + a)} + \dfrac{c}{4(c + b)} \geq \dfrac{a}{2a + b + c} + \dfrac{b}{2b + c +a} + \dfrac{c}{2c + a + b}$
$\Leftrightarrow \left[\dfrac{a}{4(a + b)} + \dfrac{b}{4(a + b)}\right] + \left[\dfrac{a}{4(a + c)} + \dfrac{c}{4(a + c)}\right] + \left[\dfrac{b}{4(b + c)} + \dfrac{c}{4(b + c)}\right] \geq \dfrac{a}{2a + b + c} + \dfrac{b}{2b + c +a} + \dfrac{c}{2c + a + b}$
$\Leftrightarrow \dfrac{1}{4} + \dfrac{1}{4} + \dfrac{1}{4} \geq \dfrac{a}{2a + b + c} + \dfrac{b}{2b + c +a} + \dfrac{c}{2c + a + b}$
$\Leftrightarrow \dfrac{3}{4} \geq \dfrac{a}{2a + b + c} + \dfrac{b}{2b + c +a} + \dfrac{c}{2c + a + b}$
Dấu = xảy ra $\Leftrightarrow a = b = c$
