Ta có ${{\log }_{2}}x+m\ge x,\forall x\in \left[ 1;3 \right]$ khi và chỉ khi $m\ge x-{{\log }_{2}}x=f\left( x \right),\forall x\in \left[ 1;3 \right]$, hay $m\ge \underset{\left[ 1;3 \right]}{\mathop{\max }}\,f\left( x \right)$

Ta tính được $f'\left( x \right)=1-\frac{1}{x\ln 2}$.

Khi đó $\left\{ \begin{align}  & f'\left( x \right)=0 \\  & x\in \left[ 1;3 \right] \\ \end{align} \right.$.$\Leftrightarrow \left\{ \begin{align}  & x=\frac{1}{\ln 2} \\  & x\in \left[ 1;3 \right] \\ \end{align} \right.\Leftrightarrow x=\frac{1}{\ln 2}$

Ta có

$f\left( 1 \right)=1,f\left( \frac{1}{\ln 2} \right)=\frac{1}{\ln 2}+{{\log }_{2}}\left( \ln 2 \right),f\left( 3 \right)=3-{{\log }_{2}}3$Vậy $\underset{\left[ 1;3 \right]}{\mathop{\max }}\,f\left( x \right)=f\left( 3 \right)=3-{{\log }_{2}}3$,  hay $m\ge 3-{{\log }_{2}}3$.