Ta có:
$\dfrac{xy}{z+1}=\dfrac{xy}{x+y+z+z}\le \dfrac{1}{4}\left(\dfrac{xy}{x+z}+\dfrac{xy}{y+z}\right)$
$\dfrac{yz}{x+1}=\dfrac{yz}{x+x+y+z}\le \dfrac{1}{4}\left(\dfrac{yz}{x+y}+\dfrac{yz}{x+z}\right)$
$\dfrac{xz}{y+1}=\dfrac{xz}{x+y+y+z}\le \dfrac{1}{4}\left(\dfrac{xz}{x+y}+\dfrac{xz}{y+z}\right)$
Suy ra:
$\dfrac{xy}{z+1}+\dfrac{yz}{x+1}+\dfrac{xz}{y+1}\le\dfrac{1}{4}(x+y+z)=\dfrac{1}{4}$
Dấu bằng xảy ra khi $x=y=z=\dfrac{1}{3}$