Áp dụng BĐT: $\dfrac{4}{x+y}\le\dfrac{1}{x}+\dfrac{1}{y}$
Ta có:
$\dfrac{ab}{a+b+2c}\le\dfrac{1}{4}\left(\dfrac{ab}{a+c}+\dfrac{ab}{b+c}\right)$
$\dfrac{bc}{b+c+2a}\le\dfrac{1}{4}\left(\dfrac{bc}{a+c}+\dfrac{bc}{a+b}\right)$
$\dfrac{ac}{c+a+2b}\le\dfrac{1}{4}\left(\dfrac{ac}{a+b}+\dfrac{ac}{b+c}\right)$
Khi đó suy ra:
$\dfrac{ab}{a+b+2c}+\dfrac{bc}{b+c+2a}+\dfrac{ac}{c+a+2b}$
$\le\dfrac{1}{4}\left(\dfrac{ab}{a+c}+\dfrac{ab}{b+c}+\dfrac{bc}{a+c}+\dfrac{bc}{a+b}+\dfrac{ac}{a+b}+\dfrac{ac}{b+c}\right)$
$=\dfrac{1}{4}\left(\dfrac{ab+bc}{a+c}+\dfrac{ab+ac}{b+c}+\dfrac{bc+ac}{a+b}\right)$
$=\dfrac{1}{4}(a+b+c)=1$
Dấu bằng xảy ra khi: $a=b=c=\dfrac{4}{3}$