Ta có $x_{n+3}-x_{n}\leq u_{n+3}-v_{n}\leq u_{n}-v_{n}=w_{n}$Bạn giải thích lại chỗ này hộ mình cái
$w_{n+3}= u_{n+3}-v_{n+3}\leq\dfrac{2}{3}w_{n}$
và $x_{n+3}-x_{n}\geq v_{n+3}-u_{n}\geq v_{n}-u_{n}=-w_{n}$
suy ra $\left | x_{n+3}-x_{n} \right |\leq w_{n}$
$\left | x_{n+4}-x_{n+3} \right |=\left | \dfrac{x_{n+1}+x_{n+2}+x_{n+3}}{3}-\dfrac{x_{n}+x_{n+1}+x_{n+2}}{3}\right |= \dfrac{\left | x_{n+3}-x_{n} \right |}{3}\leq \dfrac{1}{3}w_{n}$
tương tự: $\left | x_{n+5}-x_{n+4} \right |\leq \dfrac{1}{3}w_{n}$
$\left | x_{n+5}-x_{n+3} \right |\leq \left | x_{n+5}-x_{n+4} \right |+\left | x_{n+4}-x_{n+3} \right |\leq \dfrac{2}{3}w_{n}$
Do đó $w_{n+3}=u_{n+3}-v_{n+3}=max\left \{ x_{n+3};x_{n+4};x_{n+5} \right \}-min\left \{ x_{n+3};x_{n+4};x_{n+5} \right \}\leq \dfrac{2}{3}w_{n}.$