$\displaystyle \begin{array}{{>{\displaystyle}l}} a^{2} +b^{2} \geqslant 2ab\\ a^{2} +1\geqslant 2a\ \\ \rightarrow b^{2} +2a^{2} +3\geqslant 2( ab+a+1)\\ \rightarrow \frac{a}{2a^{2} +b^{2} +3} \leqslant \frac{a}{2( ab+a+1)} \ \\ do\ đó\ tương\ tự\ \\ 2b^{2} +c^{2} +3\geqslant 2( bc+b+1)\\ \rightarrow \frac{b}{2b^{2} +c^{2} +3} \leqslant \frac{b}{2( bc+b+1)}\\ 2c^{2} +a^{2} +3\geqslant 2( ac+c+1) \ \ \\ \rightarrow \frac{c}{2c^{2} +a^{2} +3} \leqslant \frac{c}{ \begin{array}{{>{\displaystyle}l}} 2( ac+c+1) \ \\ \end{array}}\\ \rightarrow VT\leqslant \frac{1}{2}\left(\frac{a}{ab+a+1} +\frac{b}{bc+b+1} +\frac{c}{ac+c+1}\right)\\ \rightarrow VT\leqslant \frac{1}{2}\left(\frac{a}{ab+a+abc} +\frac{b}{bc+b+abc} +\frac{c}{ac+c+abc}\right) \ \\ \rightarrow VT\leqslant \frac{1}{2}\left(\frac{1}{b+bc+1} +\frac{1}{c+ac+1} \ +\frac{1}{a+ab+1}\right)\\ Ta\ có\ :\frac{1}{b+bc+1} +\frac{1}{c+ac+1} +\frac{1}{a+ab+1}\\ =\frac{1}{b+bc+1} +\frac{b}{bc+b+1} +\frac{bc}{b+bc+1} =\frac{1+bc+1}{b+bc+1} =1\ \\ \rightarrow \ VT\leqslant \frac{1}{2} \ \\ Dấu\ =\ xảy\ ra\ khi\ a=b=c=1\ \\ Vậy\ maxA=\frac{1}{2} \ tại\ a=b=c=1\ \\ \\ \\ \\ \\ \\ \\ \end{array}$