问题
解答题
已知大于1的正数x,y,z满足x+y+z=3
(1)求证:
(2)求
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答案
(1)由柯西不等式得,
(
+x2 x+2y+3z
+y2 y+2z+3z
)[(x+2y+3z)+(y+2z+3x)+(z+2x+3y)]≥(x+y+z)2=27z2 z+2x+3y
得:
+x2 x+2y+3z
+y2 y+2z+3x
≥z2 z+2x+3y
;3 2
(2)∵
+1 log3x+log3y
+1 log3y+log3z
=1 log3z+log3x
+1 log3(xy)
+1 log3(yz)
,1 log3(zx)
由柯西不等式得:(
+1 log3(xy)
+1 log3(yz)
)(log3(xy)+log3(yz)+log3(zx)),1 log3(zx)
由柯西不等式得:(
+1 log3(xy)
+1 log3(yz)
)(log3(xy)+log3(yz)+log3(zx))≥91 log3(zx)
所以,(
+1 log3(xy)
+1 log3(yz)
)≥1 log3(zx)
=9 (log3(xy)+log3(yz)+log3(zx))
,9 2log3(xyz)
又∵3
=x+y+z≥33
.3 xyz
∴xyz≤3
.3
∴log3xyz≤
.得3 2
≥9 2log3xyz
×9 2
=32 3
所以,
+1 log3x+log3y
+1 log3y+log3z
≥3当且仅当x=y=z=1 log3z+log3x
时,等号成立.3
故所求的最小值是3.