已知大于1的正数x,y,z满足x+y+z=33.(1)求证:x2x+2y+3z+y2y+2z+3x+z2z+2x+3y≥32.(2)求1log3x+log3y+1log3y+log3z+1log3z+

题目简介

已知大于1的正数x,y,z满足x+y+z=33.(1)求证:x2x+2y+3z+y2y+2z+3x+z2z+2x+3y≥32.(2)求1log3x+log3y+1log3y+log3z+1log3z+

题目详情

已知大于1的正数x,y,z满足x+y+z=3
3

(1)求证:
x2
x+2y+3z
+
y2
y+2z+3x
+
z2
z+2x+3y
3
2

(2)求
1
log3x+log3y
+
1
log3y+log3z
+
1
log3z+log3x
的最小值.
题型:解答题难度:中档来源:不详

答案

(1)由柯西不等式得,
x2
x+2y+3z
+
y2
y+2z+3z
+
z2
z+2x+3y
)[(x+2y+3z)+(y+2z+3x)+(z+2x+3y)]≥(x+y+z)2=27
得:
x2
x+2y+3z
+
y2
y+2z+3x
+
z2
z+2x+3y
3
2

(2)∵class="stub"1
log3x+log3y
+class="stub"1
log3y+log3z
+class="stub"1
log3z+log3x
=class="stub"1
log3(xy)
+class="stub"1
log3(yz)
+class="stub"1
log3(zx)

由柯西不等式得:(class="stub"1
log3(xy)
+class="stub"1
log3(yz)
+class="stub"1
log3(zx)
)(log3(xy)+log3(yz)+log3(zx)),
由柯西不等式得:(class="stub"1
log3(xy)
+class="stub"1
log3(yz)
+class="stub"1
log3(zx)
)(log3(xy)+log3(yz)+log3(zx))≥9
所以,(class="stub"1
log3(xy)
+class="stub"1
log3(yz)
+class="stub"1
log3(zx)
)≥class="stub"9
(log3(xy)+log3(yz)+log3(zx))
=class="stub"9
2log3(xyz)

又∵3
3
=x+y+z≥3
3xyz

xyz≤3
3

log3xyz≤class="stub"3
2
.得class="stub"9
2log3xyz
≥class="stub"9
2
×class="stub"2
3
=3

所以,class="stub"1
log3x+log3y
+class="stub"1
log3y+log3z
+class="stub"1
log3z+log3x
≥3
当且仅当x=y=z=
3
时,等号成立.
故所求的最小值是3.

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