từ giải thiết, ta có $x(3x-2012)+y(3y-2012)+z(3z-2012)\leqslant 2013<=>3(x^{2}+y^{2}+z^{2})\leqslant 2013+2012(x+y+z)$Mà $3(x^{2}+y^{2}+x^{2})=(1^{2}+1^{2} + 1^{2})(x^{2}+y^{2}+z^{2})\geqslant (x+y+z)^{2}$$=>(x+y+z)^{2}\leqslant 2013+2012(x+y+z)$$<=>(x+y+z)^{2}-2012(x+y+z)-2013\leqslant 0$$<=>0\leqslant x+y+z\leqslant 2013$ $(x,y,z $ dương $)$$A=x+y+z-(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})$$\leqslant x+y+z-\frac{9}{x+y+z}(do \frac{1}{x}+\frac{1}{y}+\frac{1}{z}\geqslant \frac{9}{x+y+z})$đặt $t=x+y+z, A=t-\frac{9}{t}=f(t)(0< t\leqslant 2013)$Ta có $f(t)=1+\frac{9}{t^{2}}>0\forall t \in (0;2013]$$f(t)max=f(2013)=2013-\frac{9}{2013}=\frac{4052160}{2013}$ dấu $"="$ xảy ra $<=> x=y=z=\frac{2013}{3}$Vậy $max A=\frac{4052160}{2013}<=>x=y=z=\frac{2013}{3}$
từ giải thiết, ta có $x(3x-2012)+y(3y-2012)+z(3z-2012)\leqslant 2013<=>3(x^{2}+y^{2}+z^{2})\leqslant 2013+2012(x+y+z)$Mà $3(x^{2}+y^{2}+x^{2})=(1^{2}+1^{2} + 1^{2})(x^{2}+y^{2}+z^{2})\geqslant (x+y+z)^{2}$$=>(x+y+z)^{2}\leqslant 2013+2012(x+y+z)$$<=>(x+y+z)^{2}-2012(x+y+z)-2013\leqslant 0$$<=>0\leqslant x+y+z\leqslant 2013$ $(x,y,z $ dương $)$$A=x+y+z-(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})$$\leqslant x+y+z-\frac{9}{x+y+z}(do \frac{1}{x}+\frac{1}{y}+\frac{1}{z}\geqslant \frac{9}{x+y+z})$đặt $t=x+y+z, A=t-\frac{9}{t}=f(t)(0< t\leqslant 2013)$Ta có $f(t)=1+\frac{9}{t^{2}}>0\forall t \in (0;2013]$$f(t)max=f(2013)=2013-\frac{9}{2013}=\frac{4052160}{2013}$ dấu $"="$ xảy ra $<=> x=y=z=\frac{2013}{3}$Vậy $max A=\frac{4052160}{2013}<=>x=y=z=\frac{2013}{3}$