已知非负实数 x、y、zx、y、zx、y、z 满足 x2+y2+z2+x+2y+3z=134.x^2+y^2+z^2+x+2y+3z=\frac{13}4.x2+y2+z2+x+2y+3z=413.
(1)求 x+y+zx+y+zx+y+z 的最大值；
(2)证明： x+y+z≥22−32x+y+z\geq\frac{\sqrt{22}-3}2x+y+z≥222−3
(i)解\qquad由已知等式得 (x+12)2+(y+1)2+(z+32)2=274.(x+\frac12)^2+(y+1)^2+(z+\frac32)^2=\frac{27}4.(x+21)2+(y+1)2+(z+23)2=427.
则
[(x+12)+(y+1)+(z+32)]2≤3[(x+12)2+(y+1)2+(z+32)2]=814.[(x+\frac12)+(y+1)+(z+\frac32)]^2\leq3[(x+\frac12)^2+(y+1)^2+(z+\frac32)^2]=\frac{81}4.[(x+21)+(y+1)+(z+23)]2≤3[(x+21)2+(y+1)2+(z+23)2]=481.
故 x+y+z≤32,x+y+z\leq\frac32,x+y+z≤23,当且仅当 x=1,y=12,z=0x=1,y=\frac12,z=0x=1,y=21,z=0 时，上式等号成立.
(ii)证明\qquad因为
(x+y+z)2≥x2+y2+z2(x+y+z)^2\geq x^2+y^2+z^2(x+y+z)2≥x2+y2+z2
3(x+y+z)≥x+2y+3z3(x+y+z)\geq x+2y+3z3(x+y+z)≥x+2y+3z
所以，
(x+y+z)2+3(x+y+z)≥134.(x+y+z)^2+3(x+y+z)\geq\frac{13}4.(x+y+z)2+3(x+y+z)≥413.
解得 x+y+z≥22−32x+y+z\geq\frac{\sqrt{22}-3}2x+y+z≥222−3.
当且仅当 x=0,y=0,z=22−32x=0,y=0,z=\frac{\sqrt{22}-3}2x=0,y=0,z=222−3 时，上式等号成立.
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