Ta có $\exists x\in \left[ 1;3 \right]:{{\log }_{2}}x+m\ge \frac{1}{2}{{x}^{2}}$ khi và chỉ khi $m\ge \underset{\left[ 1;3 \right]}{\mathop{\min }}\,f\left( x \right)$, với $f\left( x \right)=\frac{1}{2}{{x}^{2}}-{{\log }_{2}}x$.

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

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

Ta có

$f\left( 1 \right)=\frac{1}{2},f\left( \sqrt{\frac{1}{\ln 2}} \right)=\frac{1}{2\ln 2}+\frac{1}{2}{{\log }_{2}}\left( \ln 2 \right)$,

$f\left( 3 \right)=\frac{9}{2}-{{\log }_{2}}3$

Vậy $\underset{\left[ 1;3 \right]}{\mathop{\min }}\,f\left( x \right)=f\left( \sqrt{\frac{1}{\ln 2}} \right)=\frac{1}{2\ln 2}+\frac{1}{2}{{\log }_{2}}\left( \ln 2 \right)$,

hay $m\ge \frac{1}{2\ln 2}+\frac{1}{2}{{\log }_{2}}\left( \ln 2 \right)$.