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(1)设x≥1,y≥1,证明x+y+1xy≤1x+1y+xy;(2)设1<a≤b≤c,证明logab+logbc+logca≤logba+logcb+logac.
题目详情
(1)设x≥1,y≥1,证明x+y+
≤
+
+xy;
(2)设1<a≤b≤c,证明logab+logbc+logca≤logba+logcb+logac.
1 |
xy |
1 |
x |
1 |
y |
(2)设1<a≤b≤c,证明logab+logbc+logca≤logba+logcb+logac.
▼优质解答
答案和解析
证明:(1)∵x≥1,y≥1,
∴x+y+
≤
+
+xy⇔xy(x+y)+1≤x+y+x2y2.
⇔(x+y)(xy-1)+(1-x2y2)≤0,
⇔(xy-1)(x+y-1-xy)≤0,
⇔(xy-1)(x+1)(1-y)≤0.
∵x≥1,y≥1,
∴xy-1≥0,x+1>0,1-y≤0,
∴(xy-1)(x+1)(1-y)≤0成立.,
∴x+y+
≤
+
+xy.
(2)设logab=x,logbc=y,则logac=xy,logca=
,logba=
,logcb=
.
∴logab+logbc+logca≤logba+logcb+logac⇔x+y+
≤
+
+xy.
∵1<a≤b≤c,∴logab≥1,logbc≥1,即x≥1,y≥1.
由(1)可知x+y+
≤
+
+xy.
∴logab+logbc+logca≤logba+logcb+logac.
∴x+y+
1 |
xy |
1 |
x |
1 |
y |
⇔(x+y)(xy-1)+(1-x2y2)≤0,
⇔(xy-1)(x+y-1-xy)≤0,
⇔(xy-1)(x+1)(1-y)≤0.
∵x≥1,y≥1,
∴xy-1≥0,x+1>0,1-y≤0,
∴(xy-1)(x+1)(1-y)≤0成立.,
∴x+y+
1 |
xy |
1 |
x |
1 |
y |
(2)设logab=x,logbc=y,则logac=xy,logca=
1 |
xy |
1 |
x |
1 |
y |
∴logab+logbc+logca≤logba+logcb+logac⇔x+y+
1 |
xy |
1 |
x |
1 |
y |
∵1<a≤b≤c,∴logab≥1,logbc≥1,即x≥1,y≥1.
由(1)可知x+y+
1 |
xy |
1 |
x |
1 |
y |
∴logab+logbc+logca≤logba+logcb+logac.
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