từ giải thiết, ta có x(3x−2012)+y(3y−2012)+z(3z−2012)⩽2013<=>3(x2+y2+z2)⩽2013+2012(x+y+z)Mà 3(x2+y2+x2)=(12+12+12)(x2+y2+z2)⩾(x+y+z)2=>(x+y+z)2⩽2013+2012(x+y+z)<=>(x+y+z)2−2012(x+y+z)−2013⩽0<=>0⩽x+y+z⩽2013 (x,y,z dương )A=x+y+z−(1x+1y+1z)⩽x+y+z−9x+y+z(do1x+1y+1z⩾9x+y+z)đặt t=x+y+z,A=t−9t=f(t)(0<t⩽2013)Ta có f(t)=1+9t2>0∀t∈(0;2013]f(t)max=f(2013)=2013−92013=40521602013 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}