Ta có:
$\quad \dfrac{ab}{4b + 4c + a}$
$= \dfrac{ab}{(2b + c) + (2b + c) + (2c + a)}$
$= \dfrac{ab}{9}\cdot \dfrac{9}{(2b + c) + (2b + c) + (2c + a)}$
$\leqslant \dfrac{ab}{9}\cdot\left(\dfrac{1}{2b +c} +\dfrac{1}{2b + c} +\dfrac{1}{2c+a}\right)\quad (BDT\ Cauchy-Schwarz)$
$=\dfrac{ab}{9}\cdot\left(\dfrac{2}{2b+c}+\dfrac{1}{2c+a}\right)$
$=\dfrac{2ab}{9(2b+c)} + \dfrac{ab}{9(2c+a)}$
Do đó:
$\dfrac{ab}{4b + 4c + a}\leqslant \dfrac{2ab}{9(2b+c)} + \dfrac{ab}{9(2c+a)}$
Chứng minh tương tự, ta được:
$\dfrac{bc}{4c + 4a + b}\leqslant\dfrac{2bc}{9(2c+a)} + \dfrac{bc}{9(2a+b)}$
$\dfrac{ca}{4a + 4b + c}\leqslant \dfrac{2ca}{9(2a+b)} + \dfrac{ca}{9(2b+c)}$
Cộng vế theo vế ta được:
$\dfrac{ab}{4b + 4c + a}+\dfrac{ca}{4a + 4b + c}+\dfrac{ca}{4a + 4b + c}\leqslant \dfrac{2ab}{9(2b+c)} + \dfrac{ab}{9(2c+a)}+ \dfrac{2bc}{9(2c+a)} + \dfrac{bc}{9(2a+b)}+ \dfrac{2ca}{9(2a+b)} + \dfrac{ca}{9(2b+c)}$
$\Leftrightarrow \dfrac{ab}{4b + 4c + a}+\dfrac{ca}{4a + 4b + c}+\dfrac{ca}{4a + 4b + c}\leqslant \dfrac{2ab+ca}{9(2b+c)} + \dfrac{2bc + ab}{9(2c + a)} +\dfrac{2ca + bc}{9(2a + b)}$
$\Leftrightarrow \dfrac{ab}{4b + 4c + a}+\dfrac{ca}{4a + 4b + c}+\dfrac{ca}{4a + 4b + c}\leqslant \dfrac{a}{9} +\dfrac{b}{9} +\dfrac{c}{9}$
$\Leftrightarrow \dfrac{ab}{4b + 4c + a}+\dfrac{ca}{4a + 4b + c}+\dfrac{ca}{4a + 4b + c}\leqslant \dfrac{a+b+c}{9}$
Dấu $=$ xảy ra $\Leftrightarrow a = b = c$
