分析法典型例题
1.已知a>b>0ab<a-b. a-b<a-b <========= (a-b)2<a-b <========= a-2ab+b<a-b < ========= b<ab < ========= b<a <========= b<a<0(已知条件).
11
2.设x,y∈R+,且x+y=1,求证:1+1≥9.
xy11
证明: 1+1+≥9 <=========
xy11 1+1+=========
x1-x≥9 <
x+12-x
·9 <========= x1-x
2+x-x2
9 <=========
x-x2+x-x2≥9x-9x2<=========
8x2-8x+2≥0 <========= (2x-1)2≥0 此式明显成立.
11
3.已知0<x<,求证:y-y2
yx+111y
证明: <=========
x+11y+1
+1y
y
y-y2 <========= y+1
1
1-y <========= y+1
1>1-y2 <========= y2>0 <========= y>0 (已知条件).
4.已知a,b,c是不全相等的正数,求证:lg 证明:lg
a+bc+bc+a
+lg>lga+lgb+lgc. 222
a+bc+bc+a
lg+lglga+lgb+lgc <========= 222a+bc+bc+b========= 2 22>lg(abc) <a+bc+bc+b
· >abc <========= 222
lg
a+ba+cb+cab,ac,bc (三式等号不能同时成立) < ========= 222a,b,c为不全相等的正数 (已知) .
4
5.已知实数a,b,c满足c<b<a,a+b+c=1,a2+b2+c2=1,求证:1<a+b<.
3
证明:∵ a+b+c=1,
14
∴ 1<a+b<⇔-<c<0. 33又∵ a2+b2+c2=1,
(a+b)2-(a2+b2)(1-c)2-(1-c2)2 ∴ ab=c-c
22而 a+b=1-c,
∴ a,b是二次方程x2-(1-c)x+c2-c=0的两个不等实根, 1
从而,△=(1-c)2-4(c2-c)>0,解得 -<c<1.
3 又∵ c<b<a,
∴ (c-a)(c-b)>0,即c2-c(a+b)+ab=c2-c(1-c)+c2-c=3c2-2c>0, 2
∴ c<0或c>(舍去).
3
14
∴ -c<0,即1<a+b<
33
xyxy
6.是否存在常数c,使得不等式c≤对任意正数x,y恒成立?
2x+yx+2yx+2y2x+y
222
解:令x=y=1,得 ≤c≤,∴ c= .
333
下面给出证明:
xy2
<========= 2x+yx+2y3
3x(x+2y)+3y(2x+y)≤2(x+2y)(2x+y) <========= x2+y2≥2xy (这个明显成立).
xy
2 <========= 3x+2y2x+y
2(x+2y)(2x+y)≤ 3x(2x+y)+3y(x+2y) <========= 2xy≤x2+y2 (这个明显成立). 2 综上所述,c=.
3
7.已知a,b,c∈R+,且a+b+c=1,求证:3a+2+3b+23c+2≤3.
证明:3a+23b+2+3c+2≤33 <=========
(
223a+2+3b+23c+2) ≤(3) <=========
3a+2+3b+2+3c+2+23a+2 3b+23b+3c+23a+3c+2≤27 <=========
3a+2 3b+2+3b+3c+2+3a+3c+2≤18. <========= 3a+2 3b+2+3b+3c+2+3a+3c+2≤6(a+b+c)+12,且a+b+c=1 <=========
3a+23b+2≤(3a+2)+(3b+2) ,且 3a+23c+2≤(3a+2)+(3c+2) ,且
3b+23c+2≤(3b+2)+(3c+2). <========= a,b,c∈R+.
8.已知a>0,b>0,2c>a+b,求证:c-c-ab<a<cc-ab. 证明 :cc-ab<a<c+c-ab <=========
c-ab<a-cc-ab < =========
|a-c|<
c-ab <=========
(a-c)2<c2-ab <========= a2-2ac+c2<c2-ab <=========
2ac>a2+ab <========= 2c>a+b(已知).
9.已知a,b,c∈R+,且ab+bc+ca=1,(1)求证:a+b+c3;
证明 a+b+c≥3<========= (a+b+c)2≥3,且a,b,c∈R+<====== === a2+b2+c2+2(ab+bc+ac)≥3
<========= a2+b2+c2≥1=ab+bc+ca <========= a2+b2≥2ab,b2+c2≥2bc,a2+c2≥2ac.
(2) 证明 ∵
bcbc
acac
3(+b+c). ab
a+b+c≥ <========= ababcabc
1
≥abc <========= abc
abc+ac+ab≤1=ab+bc+ca <=========
ab+acab+bccb+ac a=·≤·≤且·≤.
222
10.已知a,b,c,d都大于1,且loga(bcd)≤9,求证:logba+logca+logda≥1. 证明 logba+logca+logda≥1 <=========
3
111
+≥1 <========= logablogaclogad
3
≥1 <========= logab·logac·logad≤27 <========= logab·logac·logad
logab+logac+logad3log(bcd)393
========= 3 =3 ≤3 =27 <
logab·logac·logad≤
a,b,c,d都大于1,且loga(bcd)≤9.
x1+x2ππ1
11.已知函数f(x)=tanx,x∈0,,若x1,x2∈0,,且x1≠x2,求证:[f(x1)+f(x2)]>f2222. x1+x2x1+x211 证明:f(x1)+f(x2)]>f <========= tanx1+tanx2)>tan========= 2222 <
x1+x2x1+x2x1+x2
2sin·cos2221sinxsinx1sinxcosx+sinxcosx> <========= > <========= 2cosx1cosx22cosx2cosx2x1+x22x1+x2coscos
22
sin
sin(x1+x2)1sin(x1+x2)π
·> <========= 1+cos(x1+x2)>2cosx1cosx2 ,且x1,x2∈0, <======= 2cosx2cosx21+cos(x1+x2)2=1+cosx1cosx2-sinx1sinx2>2cosx1cosx2 <========= 1>cosx1cosx2+sinx1sinx2 <========= 1>cos(x1-x2) <========= x1≠x2.
12.求证:对任意实数x,不等式 证明
x≤1恒成立.
2+cosx
223sinx≤1 <
========= |3sinx|≤|2+cosx| <========= |3sinx| ≤|2+cosx| <========= 2+cosx
3sin2x≤4+4cosx+cos2x <========= 4cos2x+4cosx+1≥0 <========= (2cosx+1)2≥0 恒成立.
13.已知三角形三边长为a,b,c,面积为S,求证:a2+b2+c2≥3S. 1
证明 a2+b2+c2≥43S <========= a2+b2+c2≥3sinC <========= 2
1
a2+b2+c2≥431-cosC <========= a4+b4+c4+2a2b2+2a2c2+2b2c2≥12a2b2(1-cos2C) 2 <========= a4+b4+c4-10a2b2+2a2c2+2b2c2≥-12a2b2cos2C <========= a+b-c2 a+b+c-10ab+2ac+2bc≥-12ab========= 2ab <
4
4
4
22
22
22
22
2
2
2
a+b-c2 a+b+c-10ab+2ac+2bc≥-12ab========= 2ab <
4
4
4
22
22
22
22
222
a4+b4+c4-10a2b2+2a2c2+2b2c2≥-3(a4+b4+c4+2a2b2-2a2c2-2b2c2) <========= 2a4+2b4+2c4-2a2b2-2a2c2-2b2c2≥0 <========= (a2-b2)2+(b2-c2)2+(a2-c2)≥0.
(a-b)2a+b(a-b)214.已知a>b>0,求证:<ab<
8a28b
(a-b)2a+b(a-b)2(a-b)2a+b-2ab(a-b)2
证明 ab <========= 8a<========= 8a28b28b
2
2(a-b)22
(a-b)(a-b)2 (a-b)2(a-b)
<========= 4a<(a-b) <4b <========= 8a28b
a-ba-ba+bab
-< <======== 1ab>0 <========= 22b2ab
b1
1<a2
11
<========= b22
b11
a22
a <========= b
b
<1<a
a <========= b
11
22
b<a <========= a>b>0.
a4+b4+c4abacbc15.已知a,b,c∈R,求证:
abcc ba
a4+b4+c4abacbca4+b4+c4a2b2+a2c2+b2c2
证明 ++ <========= abc <========= abcc baabc
+
2a4+2b4+2c4-2a2b2-2a2c2-2b2c2≥0 <========= (a2-b2)2+(b2-c2)2+(a2-c2)≥0.
16.已知|a|<1,|b|<1,求证:
a+b<1.
1+ab
a+b 证明 ========= |a+b|<|1+ab| <========= |a+b|2<|1+ab|2 <========= 1+ab<1 <
a2+2ab+b2<1+2ab+a2b2 <========= a2b2-a2-b2+1>0 <========= (1-a2)(1-b2)>0 <========= 1-a2>0且1-b2>0 (或1-a2<0且1-b2<0) <======== |a|<1,|b|<1.
17.已知p,q∈(0,+∞),且p3+q3=2,求证:p+q≤2.
证明 p+q≤2 <========= (p+q)3≤23=8=2×4 <========= (p+q)3≤(p3+q3)×4 <========= p3+3p2q+3pq2+q3≤4p3+4q3 <========= p3+q3-pq(p+q)≥0 <=========
(p+q)(p2-pq+q2)-pq(p+q) ≥0 <========= (p+q)(p-q)2≥0 <========= p,q∈(0,+∞).
18.已知x∈(0,+∞),求证:x+ 证明
1
x
1x
x+1≤23. x
+
1x
1
x1≤2<======== = x
+12-1≤2-<=========
x
22
t-t-1≤23, <========= t+≤2+t-1 <======== (t+3)≤(2t-1) <=========
3t2≤4(t2-1) <========= t2≥4 <========= t≥2 <========= t=x+
19.已知a,b,c∈R+,求证:log3(a2+b2+c2)-2log3(a+b+c)≥-1
证明 log3(a2+b2+c2)-2log3(a+b+c)≥-1 <========= log3(a2+b2+c2)+1≥2log3(a+b+c) <========= log33(a2+b2+c2)≥log3(a+b+c)2 且a,b,c∈R+<========= 3(a2+b2+c2)≥(a+b+c)2 <========= 2a2+2b2+2c2≥2ab+2bc+2ac <========= a2-2ab+b2+a2-2ac+c2+b2-2bc+c2≥0 <========= (a-b)2+(b-c)2+(a-c)2≥0.
20.在锐角三角形ABC中,求证: 证明
11tanAtanB
+<
1+tanA1+tanB1+tanA1+tanB
1
≥2 (x∈(0,+∞) ). 11tanAtanB
++ <========= 1+tanA1+tanB1+tanA1+tanB
1+tanA-11+tanB-111
+<========= 1+tanA1+tanB1+tanA1+tanB1111
+1-1 <========= 1+tanA1+tanB1+tanA1+tanB
2211
+2 <========= <1 <========= 1+tanA1+tanB1+tanA1+tanB
2+tanA+tanB <1+tanAtanB+tanA+tanB <=========
π 1<tanAtanB <========= cotA<tanB <========= tan========= 2A<tanB <
2A<B <========= 锐角三角形ABC中,A+B2
ππ
abc
21.已知a,b,c为三角形的三边长,求证:+≥3.
b+c-aa+c-ba+b-c 证明
abc
≥3 <=========
b+c-aa+c-ba+b-c
2a2b2c
+≥6 <=========
b+c-aa+c-ba+b-c
2a2b2c
+1+1+1≥9 <=========
b+c-aa+c-ba+b-ca+b+ca+b+ca+b+c
+9 <=========
b+c-aa+c-ba+b-c
111 (a+b+c)≥9 <=========
b+c-aa+c-ba+b-c
111
[(b+c-a)+(a+c-b)+(a+b-c)]+≥9 <=========
b+c-aa+c-ba+b-c3
(b+c-a)+(a+c-b)+(a+b-c)≥3 (b+c-a)·(a+c-b)·(a+b-c) 且 3111111
≥3 <=========
b+c-aa+c-ba+b-cb+c-aa+c-ba+b-c
b+c-a>0,a+c-b>0,a+b-c>0<= ======== 已知a,b,c为三角形的三边长.
117
22.已知a,b,∈R+,且a+b=1,求证:ab+.
ab4
1
证明 由a,b,∈R+,且a+b=1知 0<ab
4
4(ab)2-17ab+4117117
ab+≥<========= ab+0 <========= 0 <======== 4(ab)2-17ab+4≥0
ab4ab44ab
1<========= (4ab-1)(ab-4)≥0 <========= 0<ab44 (已证).
23.(2011年高考全国卷理科压轴题)
(1)设函数f(x)=ln(1+x)--
2x
,证明:当x>0时,f(x)>0; x+2
(2)从编号1到100的100张卡片中每次随机抽取一张,然后放回,用这种方式连续抽取20次,设抽91
得的20个号码互不相同的概率为p,证明:p< <.
10e
19
证明 (1)∵ f(x)=ln(1+x)-
2(x+2)-42x4
=ln(1+x)ln(1+x)-2+ x+2x+2x+2
(x+2)2-4(1+x)14x2
∴ f′(x)=-==>0 (x>0),
1+x(x+2)(1+x)(x+2)(1+x)(x+2) ∴ f(x)在(0,+∞)单调递增,∴f(x)>f(0)=0.
100·99·98·…·8199·98·…·81
(2)易知 p=
100100
919
先证左端不等式 p< :
10
99·98·…·[1**********]
p< <========= < <========= 99·98·97·…·81<90 <========= 10010100
19
99·98·97·…·81<90 <=========
19
19
99+98+97+…+81
99·98·97·…·81<=90. 19
91
再证右端不等式 <:
10e
91 <9<ln1 <9 <-lne2 <10 >2 <========= ln========= 19ln========= ln======== e10 e10 10919=
1
29122x1
ln1+ ><========= 由(1)知ln(1+x)>,令x=
99191x+2
+29
24.(福建省2008年高考理科压轴题)
1919
已知函数f(x)=ln(1+x)-x.(Ⅰ)求f(x)的单调区间;(Ⅱ)记f(x)在区间[0,n](n∈N*)上的a1a3…a2n-1aaa最小值为bn令an=ln(1+n)-bn,求证:aaa+…+2an+1-1
a2a4…a2n224
解(Ⅰ) 由f(x)=ln(1+x)-x得f′(x)=
-x1
1,易知当x∈(-1,0)时,f′(x)>0,f(x)单调递增;1+x1+x
当x∈(0,+∞)时,f′(x)>0,f(x)单调递减.
证明(Ⅱ) 由(Ⅰ)知,f(x)在区间[0,n] (n∈N*)上单调递减,∴f(x)的最小值为f(n)=ln(1+n)-n.
于是,an=ln(1+n)-bn=n.
a1a3…a2n-1aaa1·3·…·(2n-1)11·3
2a========= +…+2n+1-1 n+1-1<22·4a2a2a42·4·…·2na2a4…a2n
<=========
1·3·…·(2n-1)
<2n+1-2n-1 <=========
2·4·…·2n
1·3·…·(2n-1)12211
<2n+1-2n-1 <========= ,
22·4·…·2n32n+122n+12n+1+2n-12n-12n-12n-12n-12n-133551
,,…, <======== <========= <========= 42n2n2n5672n+12n+12n+1(2n+1)(2n-1)<2n <========= (2n)-1<2n 显然成立.
分析法典型例题
1.已知a>b>0ab<a-b. a-b<a-b <========= (a-b)2<a-b <========= a-2ab+b<a-b < ========= b<ab < ========= b<a <========= b<a<0(已知条件).
11
2.设x,y∈R+,且x+y=1,求证:1+1≥9.
xy11
证明: 1+1+≥9 <=========
xy11 1+1+=========
x1-x≥9 <
x+12-x
·9 <========= x1-x
2+x-x2
9 <=========
x-x2+x-x2≥9x-9x2<=========
8x2-8x+2≥0 <========= (2x-1)2≥0 此式明显成立.
11
3.已知0<x<,求证:y-y2
yx+111y
证明: <=========
x+11y+1
+1y
y
y-y2 <========= y+1
1
1-y <========= y+1
1>1-y2 <========= y2>0 <========= y>0 (已知条件).
4.已知a,b,c是不全相等的正数,求证:lg 证明:lg
a+bc+bc+a
+lg>lga+lgb+lgc. 222
a+bc+bc+a
lg+lglga+lgb+lgc <========= 222a+bc+bc+b========= 2 22>lg(abc) <a+bc+bc+b
· >abc <========= 222
lg
a+ba+cb+cab,ac,bc (三式等号不能同时成立) < ========= 222a,b,c为不全相等的正数 (已知) .
4
5.已知实数a,b,c满足c<b<a,a+b+c=1,a2+b2+c2=1,求证:1<a+b<.
3
证明:∵ a+b+c=1,
14
∴ 1<a+b<⇔-<c<0. 33又∵ a2+b2+c2=1,
(a+b)2-(a2+b2)(1-c)2-(1-c2)2 ∴ ab=c-c
22而 a+b=1-c,
∴ a,b是二次方程x2-(1-c)x+c2-c=0的两个不等实根, 1
从而,△=(1-c)2-4(c2-c)>0,解得 -<c<1.
3 又∵ c<b<a,
∴ (c-a)(c-b)>0,即c2-c(a+b)+ab=c2-c(1-c)+c2-c=3c2-2c>0, 2
∴ c<0或c>(舍去).
3
14
∴ -c<0,即1<a+b<
33
xyxy
6.是否存在常数c,使得不等式c≤对任意正数x,y恒成立?
2x+yx+2yx+2y2x+y
222
解:令x=y=1,得 ≤c≤,∴ c= .
333
下面给出证明:
xy2
<========= 2x+yx+2y3
3x(x+2y)+3y(2x+y)≤2(x+2y)(2x+y) <========= x2+y2≥2xy (这个明显成立).
xy
2 <========= 3x+2y2x+y
2(x+2y)(2x+y)≤ 3x(2x+y)+3y(x+2y) <========= 2xy≤x2+y2 (这个明显成立). 2 综上所述,c=.
3
7.已知a,b,c∈R+,且a+b+c=1,求证:3a+2+3b+23c+2≤3.
证明:3a+23b+2+3c+2≤33 <=========
(
223a+2+3b+23c+2) ≤(3) <=========
3a+2+3b+2+3c+2+23a+2 3b+23b+3c+23a+3c+2≤27 <=========
3a+2 3b+2+3b+3c+2+3a+3c+2≤18. <========= 3a+2 3b+2+3b+3c+2+3a+3c+2≤6(a+b+c)+12,且a+b+c=1 <=========
3a+23b+2≤(3a+2)+(3b+2) ,且 3a+23c+2≤(3a+2)+(3c+2) ,且
3b+23c+2≤(3b+2)+(3c+2). <========= a,b,c∈R+.
8.已知a>0,b>0,2c>a+b,求证:c-c-ab<a<cc-ab. 证明 :cc-ab<a<c+c-ab <=========
c-ab<a-cc-ab < =========
|a-c|<
c-ab <=========
(a-c)2<c2-ab <========= a2-2ac+c2<c2-ab <=========
2ac>a2+ab <========= 2c>a+b(已知).
9.已知a,b,c∈R+,且ab+bc+ca=1,(1)求证:a+b+c3;
证明 a+b+c≥3<========= (a+b+c)2≥3,且a,b,c∈R+<====== === a2+b2+c2+2(ab+bc+ac)≥3
<========= a2+b2+c2≥1=ab+bc+ca <========= a2+b2≥2ab,b2+c2≥2bc,a2+c2≥2ac.
(2) 证明 ∵
bcbc
acac
3(+b+c). ab
a+b+c≥ <========= ababcabc
1
≥abc <========= abc
abc+ac+ab≤1=ab+bc+ca <=========
ab+acab+bccb+ac a=·≤·≤且·≤.
222
10.已知a,b,c,d都大于1,且loga(bcd)≤9,求证:logba+logca+logda≥1. 证明 logba+logca+logda≥1 <=========
3
111
+≥1 <========= logablogaclogad
3
≥1 <========= logab·logac·logad≤27 <========= logab·logac·logad
logab+logac+logad3log(bcd)393
========= 3 =3 ≤3 =27 <
logab·logac·logad≤
a,b,c,d都大于1,且loga(bcd)≤9.
x1+x2ππ1
11.已知函数f(x)=tanx,x∈0,,若x1,x2∈0,,且x1≠x2,求证:[f(x1)+f(x2)]>f2222. x1+x2x1+x211 证明:f(x1)+f(x2)]>f <========= tanx1+tanx2)>tan========= 2222 <
x1+x2x1+x2x1+x2
2sin·cos2221sinxsinx1sinxcosx+sinxcosx> <========= > <========= 2cosx1cosx22cosx2cosx2x1+x22x1+x2coscos
22
sin
sin(x1+x2)1sin(x1+x2)π
·> <========= 1+cos(x1+x2)>2cosx1cosx2 ,且x1,x2∈0, <======= 2cosx2cosx21+cos(x1+x2)2=1+cosx1cosx2-sinx1sinx2>2cosx1cosx2 <========= 1>cosx1cosx2+sinx1sinx2 <========= 1>cos(x1-x2) <========= x1≠x2.
12.求证:对任意实数x,不等式 证明
x≤1恒成立.
2+cosx
223sinx≤1 <
========= |3sinx|≤|2+cosx| <========= |3sinx| ≤|2+cosx| <========= 2+cosx
3sin2x≤4+4cosx+cos2x <========= 4cos2x+4cosx+1≥0 <========= (2cosx+1)2≥0 恒成立.
13.已知三角形三边长为a,b,c,面积为S,求证:a2+b2+c2≥3S. 1
证明 a2+b2+c2≥43S <========= a2+b2+c2≥3sinC <========= 2
1
a2+b2+c2≥431-cosC <========= a4+b4+c4+2a2b2+2a2c2+2b2c2≥12a2b2(1-cos2C) 2 <========= a4+b4+c4-10a2b2+2a2c2+2b2c2≥-12a2b2cos2C <========= a+b-c2 a+b+c-10ab+2ac+2bc≥-12ab========= 2ab <
4
4
4
22
22
22
22
2
2
2
a+b-c2 a+b+c-10ab+2ac+2bc≥-12ab========= 2ab <
4
4
4
22
22
22
22
222
a4+b4+c4-10a2b2+2a2c2+2b2c2≥-3(a4+b4+c4+2a2b2-2a2c2-2b2c2) <========= 2a4+2b4+2c4-2a2b2-2a2c2-2b2c2≥0 <========= (a2-b2)2+(b2-c2)2+(a2-c2)≥0.
(a-b)2a+b(a-b)214.已知a>b>0,求证:<ab<
8a28b
(a-b)2a+b(a-b)2(a-b)2a+b-2ab(a-b)2
证明 ab <========= 8a<========= 8a28b28b
2
2(a-b)22
(a-b)(a-b)2 (a-b)2(a-b)
<========= 4a<(a-b) <4b <========= 8a28b
a-ba-ba+bab
-< <======== 1ab>0 <========= 22b2ab
b1
1<a2
11
<========= b22
b11
a22
a <========= b
b
<1<a
a <========= b
11
22
b<a <========= a>b>0.
a4+b4+c4abacbc15.已知a,b,c∈R,求证:
abcc ba
a4+b4+c4abacbca4+b4+c4a2b2+a2c2+b2c2
证明 ++ <========= abc <========= abcc baabc
+
2a4+2b4+2c4-2a2b2-2a2c2-2b2c2≥0 <========= (a2-b2)2+(b2-c2)2+(a2-c2)≥0.
16.已知|a|<1,|b|<1,求证:
a+b<1.
1+ab
a+b 证明 ========= |a+b|<|1+ab| <========= |a+b|2<|1+ab|2 <========= 1+ab<1 <
a2+2ab+b2<1+2ab+a2b2 <========= a2b2-a2-b2+1>0 <========= (1-a2)(1-b2)>0 <========= 1-a2>0且1-b2>0 (或1-a2<0且1-b2<0) <======== |a|<1,|b|<1.
17.已知p,q∈(0,+∞),且p3+q3=2,求证:p+q≤2.
证明 p+q≤2 <========= (p+q)3≤23=8=2×4 <========= (p+q)3≤(p3+q3)×4 <========= p3+3p2q+3pq2+q3≤4p3+4q3 <========= p3+q3-pq(p+q)≥0 <=========
(p+q)(p2-pq+q2)-pq(p+q) ≥0 <========= (p+q)(p-q)2≥0 <========= p,q∈(0,+∞).
18.已知x∈(0,+∞),求证:x+ 证明
1
x
1x
x+1≤23. x
+
1x
1
x1≤2<======== = x
+12-1≤2-<=========
x
22
t-t-1≤23, <========= t+≤2+t-1 <======== (t+3)≤(2t-1) <=========
3t2≤4(t2-1) <========= t2≥4 <========= t≥2 <========= t=x+
19.已知a,b,c∈R+,求证:log3(a2+b2+c2)-2log3(a+b+c)≥-1
证明 log3(a2+b2+c2)-2log3(a+b+c)≥-1 <========= log3(a2+b2+c2)+1≥2log3(a+b+c) <========= log33(a2+b2+c2)≥log3(a+b+c)2 且a,b,c∈R+<========= 3(a2+b2+c2)≥(a+b+c)2 <========= 2a2+2b2+2c2≥2ab+2bc+2ac <========= a2-2ab+b2+a2-2ac+c2+b2-2bc+c2≥0 <========= (a-b)2+(b-c)2+(a-c)2≥0.
20.在锐角三角形ABC中,求证: 证明
11tanAtanB
+<
1+tanA1+tanB1+tanA1+tanB
1
≥2 (x∈(0,+∞) ). 11tanAtanB
++ <========= 1+tanA1+tanB1+tanA1+tanB
1+tanA-11+tanB-111
+<========= 1+tanA1+tanB1+tanA1+tanB1111
+1-1 <========= 1+tanA1+tanB1+tanA1+tanB
2211
+2 <========= <1 <========= 1+tanA1+tanB1+tanA1+tanB
2+tanA+tanB <1+tanAtanB+tanA+tanB <=========
π 1<tanAtanB <========= cotA<tanB <========= tan========= 2A<tanB <
2A<B <========= 锐角三角形ABC中,A+B2
ππ
abc
21.已知a,b,c为三角形的三边长,求证:+≥3.
b+c-aa+c-ba+b-c 证明
abc
≥3 <=========
b+c-aa+c-ba+b-c
2a2b2c
+≥6 <=========
b+c-aa+c-ba+b-c
2a2b2c
+1+1+1≥9 <=========
b+c-aa+c-ba+b-ca+b+ca+b+ca+b+c
+9 <=========
b+c-aa+c-ba+b-c
111 (a+b+c)≥9 <=========
b+c-aa+c-ba+b-c
111
[(b+c-a)+(a+c-b)+(a+b-c)]+≥9 <=========
b+c-aa+c-ba+b-c3
(b+c-a)+(a+c-b)+(a+b-c)≥3 (b+c-a)·(a+c-b)·(a+b-c) 且 3111111
≥3 <=========
b+c-aa+c-ba+b-cb+c-aa+c-ba+b-c
b+c-a>0,a+c-b>0,a+b-c>0<= ======== 已知a,b,c为三角形的三边长.
117
22.已知a,b,∈R+,且a+b=1,求证:ab+.
ab4
1
证明 由a,b,∈R+,且a+b=1知 0<ab
4
4(ab)2-17ab+4117117
ab+≥<========= ab+0 <========= 0 <======== 4(ab)2-17ab+4≥0
ab4ab44ab
1<========= (4ab-1)(ab-4)≥0 <========= 0<ab44 (已证).
23.(2011年高考全国卷理科压轴题)
(1)设函数f(x)=ln(1+x)--
2x
,证明:当x>0时,f(x)>0; x+2
(2)从编号1到100的100张卡片中每次随机抽取一张,然后放回,用这种方式连续抽取20次,设抽91
得的20个号码互不相同的概率为p,证明:p< <.
10e
19
证明 (1)∵ f(x)=ln(1+x)-
2(x+2)-42x4
=ln(1+x)ln(1+x)-2+ x+2x+2x+2
(x+2)2-4(1+x)14x2
∴ f′(x)=-==>0 (x>0),
1+x(x+2)(1+x)(x+2)(1+x)(x+2) ∴ f(x)在(0,+∞)单调递增,∴f(x)>f(0)=0.
100·99·98·…·8199·98·…·81
(2)易知 p=
100100
919
先证左端不等式 p< :
10
99·98·…·[1**********]
p< <========= < <========= 99·98·97·…·81<90 <========= 10010100
19
99·98·97·…·81<90 <=========
19
19
99+98+97+…+81
99·98·97·…·81<=90. 19
91
再证右端不等式 <:
10e
91 <9<ln1 <9 <-lne2 <10 >2 <========= ln========= 19ln========= ln======== e10 e10 10919=
1
29122x1
ln1+ ><========= 由(1)知ln(1+x)>,令x=
99191x+2
+29
24.(福建省2008年高考理科压轴题)
1919
已知函数f(x)=ln(1+x)-x.(Ⅰ)求f(x)的单调区间;(Ⅱ)记f(x)在区间[0,n](n∈N*)上的a1a3…a2n-1aaa最小值为bn令an=ln(1+n)-bn,求证:aaa+…+2an+1-1
a2a4…a2n224
解(Ⅰ) 由f(x)=ln(1+x)-x得f′(x)=
-x1
1,易知当x∈(-1,0)时,f′(x)>0,f(x)单调递增;1+x1+x
当x∈(0,+∞)时,f′(x)>0,f(x)单调递减.
证明(Ⅱ) 由(Ⅰ)知,f(x)在区间[0,n] (n∈N*)上单调递减,∴f(x)的最小值为f(n)=ln(1+n)-n.
于是,an=ln(1+n)-bn=n.
a1a3…a2n-1aaa1·3·…·(2n-1)11·3
2a========= +…+2n+1-1 n+1-1<22·4a2a2a42·4·…·2na2a4…a2n
<=========
1·3·…·(2n-1)
<2n+1-2n-1 <=========
2·4·…·2n
1·3·…·(2n-1)12211
<2n+1-2n-1 <========= ,
22·4·…·2n32n+122n+12n+1+2n-12n-12n-12n-12n-12n-133551
,,…, <======== <========= <========= 42n2n2n5672n+12n+12n+1(2n+1)(2n-1)<2n <========= (2n)-1<2n 显然成立.