Inequality
1.
(a)
(m3 + 1) – (m2 + m) = (m – 1)2 (m + 1) > 0, unless m = 1, where equality holds.
∴
m3 + 1 > m2 + m.
(b)
(x5 + y5 ) – (x4y + xy4) = (x2 + y2)(x + y)(x – y)2 > 0, unless x = y, where equality holds.
∴
x5 + y5 > x4y + xy4.
(c)
(b + c – a)2 + (c + a – b)2 + (a + b – c)2 – (bc + ca + ab)
= 3(a2 + b2 + c2) – 3(ab + bc + ca)
3
= [(a − b) 2 + ( b − c) 2 + (c − a ) 2 ] > 0 , unless a = b = c, where equality holds.
2
∴
(b + c – a)2 + (c + a – b)2 + (a + b – c)2 > bc + ca + ab.
(d)
a + b > 2 ab , b + c > 2 bc , c + a > 2 ca , equalities hold iff a = b = c .
Multiply these three inequalities, we get:
(a + b)(b + c)(c + a) > 8abc.
(e)
a2b + b2c + c2a + ab2 + bc2 + ca2 – 6abc
= a(b – c)2 + b(c – a)2 + c(a – b)2 > 0, unless a = b = c , where equality holds.
∴
a2b + b2c + c2a + ab2 + bc2 + ca2 > 6abc.
(f)
a+b
=
∴
(g)
b+c
+
c+a
−6=
ab(a + b) + bc( b + c) + ac( b + a ) − 6abc
c
a
b
2
a ( b − c) + b (c − a ) 2 + c(a − b ) 2
a+b
c
+
abc
b+c
a
+
c+a
b
abc
> 0 , unless a = b = c , where equality holds.
>6.
(ay + bx ) 2 − 4(ay)( bx )
ay + bx ay + bx
⎛ x y ⎞⎛⎜ a b ⎞⎟
−
=
×
−
=
4
4
+
+
⎟
⎜
⎝ a b ⎠⎜⎝ x y ⎟⎠
abxy
ab
xy
=
∴
(h)
+
(ay − bx ) 2
abxy
> 0 , unless ay = bx, where equality holds.
x y
⎛ x y ⎞⎛⎜ a b ⎞⎟
⎜ + ⎟⎜ + ⎟ > 4 , equality holds ⇔ = .
a b
⎝ a b ⎠⎝ x y ⎠
(a + b + c )⎛⎜ + + ⎞⎟ − 9 =
1
=
∴
1
1
(a + b + c)(bc + ca + ab) − 9abc
⎝a b c⎠
a ( b − c) 2 + b (c − a ) 2 + c(a − b ) 2
abc
> 0 , unless a = b = c , where equality holds.
abc
1 1 1
(a + b + c )⎛⎜ + + ⎞⎟ > 9 .
⎝a b c⎠
(i)
In (h), replace a by l/a , b by b/m, c by c/n, result follows.
l m n
Equality holds ⇔ = = .
a b c
(j)
x7 + y7 – (x4y3 + x3y4) = (x4 – y4)(x3 – y3) = (x2 + y2)(x + y)(x – y)2(x2 + xy + y2) > 0
Equality holds ⇔ x = 1.
∴
x7 + y7 > x4y3 + x3y4,
(k)
x7 + 1 – (x6 + x) = (x6 – 1)(x – 1) = (x – 1)2(x5 + x4 + x3 + x2 + x + 1) > 0
Equality holds ⇔ x = 1.
∴
x7 + 1 > x6 + x,
1
(l)
x5 + x-5 – (x2 + x-2) = x5 – x2 – (x-2 – x-5) = x2(x3 – 1) – xP-5(x3 – 1) = (x3 – 1)(x2 – x-5)
= x-5(x3 – 1)(x7 – 1) = x-5(x – 1)2(x2 + x + 1)(x6 + x5 + x4 + x3 + x2 + x + 1) > 0
Equality holds ⇔ x = 1.
∴
x5 + x-5 > x2 + x-2,
(m)
a3 + b3 + c3 – 3abc = (a + b + c)(a2 +b2 + c2 – ab – bc – ca)
1
= (a + b + c)[(a − b) 2 + ( b − c) 2 + (c − a ) 2 ] > 0 , unless a = b = c, where equality holds.
2
∴
a3 + b3 + c3 > 3abc.
Also, (a3 + b3 + c3)3 > 27a3b3c3.
Replace a3 by a , b3 by b , c3 by c,
∴
(a + b + c)3 > 27abc.
(n)
Put
2A = x + y – z, 2B = y + z – x,
2C = z + x – y
The inequality is changed to:
(A + B)(B + C)(C + A) > 8ABC (*)
(1)
If A, B, C are all positive, (*) is proved in part (d).
(2)
If one of A, B, C is negative,
The R.H.S. of (*) is negative while the L.H.S. is positive. ∴ (*) is always true.
(3)
If two or more of the A, B, C are negative, without lost of generality,
let A, B < 0.
∴
x + y – z < 0 and y + z – x < 0.
Adding the inequalities, we have y < 0. Contradict to the given that y > 0.
2.
Since n is a positive integer, n + 1 < n + 2 < … < n + n = 2n (Equality holds when n = 1)
1
1
1
1
1
1
1 1
+
+ ... +
>
+
+ ... +
= n×
=
∴
n +1 n + 2
2n 2n 2n
2n
2n 2
3.
(1)
1
n+i
−
p
∑
1
n + i +1
1
(n + i)
i =1
⎡ 1
=⎢
+
⎣ n +1
⎡ 1
=⎢
+
⎣ n +1
1
(2)
−
n + i −1
=
( n + i )( n + i + 1)
i =1
1 ⎤ ⎡
p
<
1
(n + i) 2
1 ⎤
⎡ 1
⎢⎣ n + i − n + i + 1 ⎥⎦ =
p
∑
>
2
1
p
∑
i =1
p −1
1
n +i
−
p
∑
i =1
1
n + i +1
⎤
∑ n + i ⎥⎦ − ⎢⎣ n + p + 1 + ∑ n + i + 1⎥⎦
1
i=2
i =1
p −1
⎤ ⎡ 1
+
⎥−⎢
n + i + 1⎦ ⎣ n + p + 1
∑
i =1
1
1
n+i
=
1
( n + i − 1)( n + i )
>
1
p −1
⎤
∑ n + i + 1⎥⎦ = n + 1 − n + p + 1
1
1
1
i =1
1
( n + i) 2
1 ⎤ 1
1
⎡ 1
−
= −
⎢
⎥
(n + i)
n +i⎦ n n + p
i =1
i =1 ⎣ n + i − 1
Combining (1) and (2), result follows.
p
∑
4.
1
2
1 = (i + 1) – i =
i +1 − i =
∑[
n
(
p
∑
<
i +1 − i
1
i +1 + i
<
i =1
i+ i
n
∑
2
1
i =1
1
i
=
)
1
2 i
1
i
i =1
n
i +1 + i
1
] 1∑
2
i +1 − i <
n +1 − 1 <
)(
,
result follows.
2
5.
By Mathematical Induction.
1
Let P(s) be the proposition :
<
2 s
1
For P(2),
∴
2 s
P(2) is true.
=
1
2 2
≈ 0.354,
1
4
s
C s2 s <
1
4
s
1
2s + 1
C s2 s =
1
4
2
1
Assume P(k) is true for some k ∈ N, that is,
1
C 24 ≈ 0.375,
<
1
C 2k k <
2s + 1
1
2k + 1
2 k 4
2k + 1 ( 2k + 2)(2k + 1)
=
,
For P(k + 1), From (1), multiply throughout by
2( k + 1)
4( k + 1) 2
1 2k + 1
1 ( 2k + 2)(2k + 1) k
1
2k + 1
< k
C2k <
2
4( k + 1)
2 k 2( k + 1) 4
2k + 1 2( k + 1)
1
2k + 1
4 k ( k + 1)
<
1
4
k +1
C 2k +k1+ 2 <
k
2k + 1
2(k + 1)
=
1
5
≈ 0.447
…. (1)
…. (2)
2
But
(2k + 1) 2
4k 2 + 4k + 1 1
1
⎡ 1 2k + 1 ⎤
=
=
>
⎥
⎢
2
[4k (k + 1)] [4(k + 1)] 4k + 4k 4(k + 1) 4(k + 1)
⎣ 4 k ( k + 1) ⎦
2
⎡ 2k + 1 ⎤
2k + 1
1 ( 2 k + 3)( 2 k + 1)
1 4 k 2 + 8k + 3
1
And
=
=
=
<
⎢
⎥
2
2
2
4 k + 8k + 4 2 k + 3 4 k + 8k + 4
2 k + 3 4 k + 8k + 4 2 k + 3
⎣ 2( k + 1) ⎦
1
1
1
< k +1 C k2 +k1+ 2 <
From (2), we get:
2k + 3
2 k +1 4
∴
P(k + 1) is also true.
By the Principle of Mathematical Induction, P(s) is true s ∈ N.
6.
Method 1
θ
θ
θ ⎛
θ ⎞
θ ⎞
⎛
2 θ
− 2 cot + 1 ≥ 0 ⇒ 2 cot 2 ≥ 2 cot + ⎜ cot 2 − 1⎟
⎜ cot − 1⎟ ≥ 0 ⇒ cot
⎝
2 ⎠
2
2
2
2 ⎝
2 ⎠
θ
As 0 < θ < π, divide (1) by 2 cot ≥ 0 ,
2
θ
cot 2 − 1
θ
2
cot ≥ 1 +
= 1 + cot θ
∴
θ
2
2 cot
2
Method 2
θ
θ
1
⎛θ⎞
f ' (θ ) = − csc 2 ⎜ ⎟ + csc 2 θ
Let f (θ ) = cot 2 − 2 cot + 1,
⎝2⎠
2
2
2
2
π
π
1
1
⎛π⎞
f ' ⎜ ⎟ = − csc 2 + csc 2 = −
2 +1 = 0
⎝2⎠
2
4
2
2
θ
θ
θ
θ ⎞⎛ 1 ⎞
θ ⎞⎛
1
1⎛
f " (θ ) = − ⎜ 2 csc ⎟⎜ − csc cot ⎟⎜ ⎟ + 2 csc θ( − csc θ cot θ ) = csc 2 cot − 2 cot θ cot θ
2
2
2
2⎝
2 ⎠⎝
2
2 ⎠⎝ 2 ⎠
π
π
π
π 1
⎛π⎞ 1
f " ⎜ ⎟ = csc 2 cot − 2 cot cot = × 2 × 1 − 2 × 1× 0 = 1 > 0
⎝2⎠ 2
4
4
2
2 2
π
∴
f(θ) is the minimum when θ = .
2
θ
⎛π⎞
cot ≥ 1 + cot θ , ∀θ ∈ (0, π)
∴
f(θ) > f ⎜ ⎟ = 0 ,
⎝2⎠
2
2
...( 1)
( )
3
7.
(bc-ad)2 ≥ 0 ⇒
b2c2 + a2d2 ≥ 2abcd
⇒
b2c2 + 2abcd + a2d2 ≥ 4abcd
2
2
∴
(bc + ad) ≥ [2√(abcd)] .
As a, b, c, d > 0, taking the positive square root, we have: bc + ad ≥ 2√(abcd).
ab + bc + cd + da ≥ ab + 2√(abcd) + cd
(a + c )( b + d ) ≥ ( ab + cd )
2
(a + c )( b + d ) ≥ ab + cd
Take the positive square root again,
8.
9.
a 3 + b3
2
(a)
∴
3
2
⎛a+b⎞
−⎜
⎟ = (a + b )(a − b ) ≥ 0 .
⎝ 2 ⎠
8
3
Let d be the common difference of the arithmetic progression.
For 0 ≤ k ≤ n,
ak+1 an-k = (a1 + kd)[a + (n – k – 1) d]
= a12 + (n – 1)a1d + k (n – k – 1) d2
≥ a12 + (n – 1)a1d
= a1 [a1 + (n – 1) d]
= a1an
n −1
Πa
≥
a
k +1 n − k
k =0
∴
Πa a
10.
k +1
≥ (a 1a n )
n −k
⎛
⇒⎜
⎝
n
k =0
a 1 + a n = a k +1 + a n − k ≥ 2
n −1
Π (a
1
+ an ) ≥
n
n −1
Π2
a 1a 2 ....a n ≤
ak = k,
⇒ (a 1 + a n ) ≥ 2
n
a k +1a n −k
n
k =0
n
Π
k =1
2
⎞
n
a k ⎟ ≥ (a 1a n )
⎠
⎛
⎜
⎝
a k +1 a n − k
⎞
ak ⎟
⎠
k =1
n
Π
2
a1 + a n
2
a 1a n ≤ n a 1a 2 ....a n ≤
a1 + a n
2
1× n ≤ n 1.2....n ≤
for 0 ≤ k ≤ n,
1+ n
2
∴
n < n n! <
n +1
2
.
Cauchy-Bunyakovskii-Schwarz (CBS) inequality:
n
⎡ n
⎤
⎡ n
⎤
⎡ n
⎤
( a i x + b ) 2 ≥ 0 ⇒ ( a i x + b ) 2 = ⎢ ( a i ) 2 ⎥ x 2 + 2 ⎢ ( a i b i )⎥ x + ⎢ ( b i ) 2 ⎥ ≥ 0
⎣ i=1
⎦
⎣ i=1
⎦
⎣ i=1
⎦
i =1
∑
∴
11.
Πa Πa
k =0
For 0 ≤ k ≤ n,
Combining (a) and (b),
Putting
n −1
n
k =0
∴
⇒
1 n
k =0
n −1
a 1a n ≤ a 1a 2 ....a n
∴
(b)
n −1
∑
∑
n
⎡ n
⎡ n
⎤
2 ⎤⎡
2⎤
Δ ≤ 0 ⇒ ⎢ (a i ) ⎥ ⎢ ( b i ) ⎥ ≥ ⎢ (a i b i )⎥
⎣ i =1
⎦ ⎣ i =1
⎦ ⎣ i=1
⎦
∑
∑
∑
∑
2
By CBS inequality : (a 1b1 + a 2 b 2 + .... + a n b n ) 2 ≤ (a 12 + a 2 2 + .... + a n 2 )(b12 + b 2 2 + .... + b n 2 )
Put
bi = 1 ∀ i = 1, 2,… , n.
2
2
2
(a 1 + a 2 + .... + a n ) 2 ≤ (a 1 + a 2 + .... + a n )(n )
Take the square root, a 1 + a 2 + .... + a n ≤ n (a 1 2 + a 2 2 + .... + a n 2 )
12.
Method 1
By CBS inequality : (a 1b1 + a 2 b 2 + .... + a n b n ) 2 ≤ (a 12 + a 2 2 + .... + a n 2 )(b12 + b 2 2 + .... + b n 2 )
Put
ai2 = xi,
bi2 = 1/xi
∀ i = 1, 2,… , n.
⎛
1
⎜ x1
+ x2
⎜
x1
⎝
∴
1
x2
2
⎛ 1
1 ⎞
1
1 ⎞
⎟ ≤ ( x 1 + x 2 + ... + x n )⎜⎜ +
⎟
+ ... +
⎟
x n ⎟⎠
xn ⎠
⎝ x1 x 2
+ ... + x n
⎛ 1
( x 1 + x 2 + .... + x n )⎜⎜
⎝ x1
+
1
x2
+ .... +
1 ⎞
⎟⎟ ≥ n 2
xn ⎠
4
Method 2
x 1 + x 2 + .... + x n ≥ n n x 1 x 2 ....x n
1
x1
+
1
x2
+ .... +
1
xn
≥ nn
1 1
x1 x 2
...
(A.M. ≥ G.M.)
1
(A.M. ≥ G.M.)
xn
⎛ 1
1
1 ⎞
⎟ ≥ n2
Multiply the two inequalities, ( x 1 + x 2 + .... + x n )⎜⎜ +
+ .... +
x n ⎟⎠
⎝ x1 x 2
Method 3
a1 ≥ a2 ≥ … ≥ an
(Tchebychef’s inequality) If
∑a ∑b ≥ ∑a b
i
n
i
n
i
i
n
.
Without lost of generality, let x1 ≥ x2 ≥ … ≥ xn and y1 ≥ y2 ≥ … ≥ yn,
and put
ai = xi, bi = 1/xi for all i = 1,2, …, n in Tchebychef’s inequality,
1
1
xi
xi
1
xi n
xi
xi
≥
= =1
⇒
≥ n2
xi
n
n
n
n
Method 1
n
n
⎛ 1⎞
⎛ n +1⎞
n
n +1
n
n +1
n > n +1
( n + 1) < n
⇔
⇔
⇔
⎟ <n
⎜1 + ⎟ < n
⎜
⎝ n⎠
⎝ n ⎠
n
⎛ 1⎞
Let P(n) be the proposition :
⎜1 + ⎟ < n .
⎝ n⎠
3
64
⎛ 1⎞
For P(3),
∴ P(3) is true.
<3
⎜1 + ⎟ =
⎝ 3⎠
27
k
⎛ 1⎞
Assume P(k) is true for some k ∈ N. i.e. ⎜1 + ⎟ < k
… (1)
⎝ k⎠
k +1
k
k
1 ⎞
1 ⎞ ⎛
1 ⎞ ⎛ 1⎞ ⎛
1 ⎞
1 ⎞
⎛
⎛
For P(k + 1), ⎛⎜1 +
1
1
1
<
+
+
=
+
⎟ , by (1)
⎟ < k ⎜1 +
⎟ ⎜1 +
⎟ ⎜
⎟ ⎜
⎜
⎟
⎝ k +1⎠
⎝ k +1⎠ ⎝ k +1 ⎠ ⎝ k ⎠ ⎝ k +1⎠
⎝ k +1 ⎠
k
=k+
< k +1 .
k +1
∴
P(k + 1) is also true.
By the Principle of Mathematical Induction, P(n) is true ∀n ∈ N.
∑ ∑
13.
b1 ≥ b2 ≥ … ≥ bn, then
and
∑
∑ ∑
Method 2
1
d
Let f ( x ) = x x ,
If x > e, then
d
dx
dx
1
xx =
1
x
1
xx
−1
1
dx
1
x + (ln x )x x
d 1
dx x
1
−2
= x x (1 − ln x )
1
xx <0.
1
14.
x x is decreasing when x is increasing and x ≥ 3.
∴
Putting x = 3, 4, …., n , n + 1 in f(x), since 3 < 4 < … < n < n + 1, we have:
3
3 > 4 4 > 5 5 > .... > n n > n +1 n + 1 > ....
Method 1
n −1
⎛ 1⎞
n −1
⇔
n > n n +1
n n > ( n + 1) n −1 ⇔
n > ⎜1 + ⎟
⎝ n⎠
Use Mathematical Induction to prove the last assertion similar to 13 (Method 1).
Method 2
1
Consider g ( x ) = ( x + 1) x , use the same method as in 13 (Method 2) to show that g(x) is decreasing.
5
15.
For the sequence of positive numbers 1, a, a2, …, an-1, apply A.M. ≥ G.M.
1
1 + a + a 2 + ... + a n −1
≥ 1.a.a ....a
n
Multiply by a – 1 > 0,
n
2
(a − 1)(1 + a + a 2 + ... + a n −1 ) ≥ na
n −1
(a − 1)
⎧ n ( n −1) ⎫ n
≥ n ⎨a 2 ⎬
⎩
⎭
1 + a + a + ... + a
⇒
n −1
⎛ n +1
⎞
a n − 1 ≥ n ⎜⎜ a 2 − a 2 ⎟⎟
⎝
⎠
n −1
2
n −1
⇒
2
16.
Let 1 ≤ r ≤ n, r ∈ N.
2 + 4 + 6 + ... + 2n n
> 2.4.6...( 2n ) , A.M. > G.M.
(n – r)(r – 1) ≥ 0 ⇒ r(n – r + 1) ≥ n.
n
Put
r = 1,
1.n
=n
1 ( 2 + 2n ) n n
Put
r = 2,
2(n – 1) > n
> 2.4.6...( 2n )
n
2
Put
r=3
3(n – 2) > n
∴
2 ⋅ 4 ⋅ 6 ⋅ …. ⋅ (2n) < (n + 1)n
:
:
:
:
:
Put
r=n
n.1
=n
Multiply the inequalities (n!)2 > nn.
17.
Similar to No. 1(m).
18.
[n – (2r – 1)]2 ≥ 0
⇒
⇒
⇒
Putting r = 1, 2, …, n.
n2 ≥ [2n – 1] [1]
n2 ≥ [2(n-1) – 1] [3]
n2 ≥ [2(n-3) – 1] [5]
:
:
:
2
n ≥ [1] [2n – 1]
∴
[n2]n ≥ [1.3.5…(2n – 1)]2.
Multiply all inequalities,
19.
n2 – 2n(2r – 1) + (2r – 1)2 ≥ 0
n2 ≥ 2n(2r – 1) + (2r – 1)2
n2 ≥ [2(n – r + 1) – 1] [2r – 1]
Method 1
Similar to the first part of Number 15 where a = 2.
Method 2
Let P(n) be the proposition : 2 n > 1 + n 2 n −1 .
For P(2),
22 = 4 =1+ 2×1.5 >1+ 2 2
For P(3),
23 = 8 = 1 + 3 ×
7
3
> 1 + 3 22
∴ P(3) is true.
2 k > 1 + k 2 k −1 … (1)
Assume P(k) is true for some k ∈ N. i.e.
For P(k + 1),
∴ P(2) is true.
(
)
2 k +1 = 2 × 2 k > 2 1 + k 2 k −1 , by (1)
= 2 + 2 k 2 k −1 > 1 + 2 k 2 k −1 = 1 + 2 k 2 k > 1 + (1 + 0.4) k 2 k
= 1 + ( k + 0.4 k ) 2 k > 1 + ( k + 1) 2 k , where k ≥ 3.
∴
P(k + 1) is also true.
By the Principle of Mathematical Induction, P(n) is true ∀n ∈ N.
20.
1 + 2 + ... + n
n
But
≥ n 1.2....n
nn ≥ n!
Multiply (1) and (2),
⇒
n ( n + 1)
2n
⎛ n +1⎞
≥ n n! ⇒ ⎜
⎟ ≥ n! ⇒
⎝ 2 ⎠
n
⎛ n +1⎞
⎜
⎟
⎝ 2 ⎠
2n
≥ ( n!) 2
(1)
(2)
⎛ n +1⎞
nn ⎜
⎟
⎝ 2 ⎠
2n
> ( n!) .
3
6
21.
(i)
(ii)
Since sum of two sides of a Δ ≥ third side, therefore (x + y – z), (y + z – x), (z + x – y) ≥ 0 .
Apply A.M. ≥ G.M.
(x + y − z ) + (y + z − x ) + (z + x − y ) ≥ 3 (x + y − z )(y + z − x )(z + x − y )
3
3
∴
(x + y + z) > 27(x + y – z)(y + z – x)(z + x – y)
(y + z − x ) + (z + x − y ) ≥ (y + z − x )(z + x − y )
2
z≥
∴
Similarly,
(y + z − x )(z + x − y )
y ≥ (x + y − z )(y + z − x )
x ≥ (x + y − z )(z + x − y )
(1) × (2) × (3),
22.
E
….
(1)
….
(2)
….
(3)
xyz > (x + y – z)(y + z – x)(z + x – y)
1! 2! 3! 4! 5! ... (2n − 1)! 1! 2! ...(n − 1)! n! (n + 1)! ... (2n − 1)!
=
(2 ×1)! (2 × 2)! ... [2(n − 1)] !
2! 4! ... [2(n − 1)] !
(n − 1)! (2n − 1)! × n! = n! nΠ−1 k! (n + k )!
1! (n + 1)! 2! (n + 2)!
×
× ... ×
=
k =1
(2 × 1)! (2 × 2)!
(2k )!
[2(n − 1)] !
= 1! 3! 5! ….(2n – 1)! =
But
k! (n + k )! ⎡ n + 1⎤ ⎡ n + 2 ⎤ ⎡ n + k ⎤
=
× n!≥ n! , since n ≥ k .
...
(2k )! ⎢⎣ k + 1⎥⎦ ⎢⎣ k + 2 ⎥⎦ ⎢⎣ k + k ⎥⎦
∴ E
≥ n! Π n!= n!(n!)
n −1
k =1
n −1
= (n!)
n
E = a2 (p – q)(p – r) + b2 (q – r)(q – p) + c2 (r – p)(r – q)
= ( b2 + c2 – 2bc cos A ) (p – q)(p – r) + b2 (q – r)(q – p) + c2 (r – p)(r – q)
= b2 [(p – q)(p – r) + (q – r)(q – p)] + c2 [(p – q)(p – r) + (r – p)(r – q) ] – 2bc (p – q)(p – r) cos A
= b2 (p – q)2 + c2 (p – r)2 – 2bc (p – q)(p – r) cos A
≥ b2 (p – q)2 + c2 (p – r)2 – 2bc (p – q)(p – r) , it is trivial that E ≥ 0 if (p – q)(p – r) cos A ≤ 0.
= [ b(p – q) – c(p – r) ]2 ≥ 0
(ii) E = a2yz + b2zx + c2xy
= a2yz + b2z (–y –z) + c2 (–y –z)y = (a2 – b2 – c2)yz – b2z2 – c2y2
2 2
= –2bc cos A yz – b z – c2y2
≤ –2bc yz – b2z2 – c2y2, it is trivial that E ≤ 0 if 2bc cos A ≥ 0
= –(bz + cy)2 ≤ 0
23.
(i)
24.
E (a, b, c) = a(a – b)(a – c) + b(b – c)(b – a) + c(c – a)(c – b)
Since E(a, b, c) = E(b, a, c) = E(b, c, a) = E(c, a, b) = … , E is cyclic and symmetric.
Therefore, without lost of generality, let a ≤ b ≤ c.
E (a, b, c)
= a(a – b) [(a – b) + (b – c)] + b(b – c)(b – a) + c(c – a)(c – b)
= a(a – b)2 + [a(a – b) (b – c) – b(b – c)(a – b)] + c(c – a)(c – b)
= a(a – b)2 + (a – b)2 (b – c) + c(c – a)(c – b) ≥ 0 , since c – a ≥ 0 , c – b ≥ 0.
Similarly let F(a, b, c) = a2(a – b)(a – c) + b2(b – c)(b – a) + c2(c – a)(c – b)
Since F(a, b, c) = F(b, a, c) = F(b, c, a) = F(c, a, b) = … , F is cyclic and symmetric.
F(a, b, c)
= a2(a – b) [(a – b) + (b – c)] + b2(b – c)(b – a) + c2(c – a)(c – b)
= a2(a – b)2 + [a2(a – b) (b – c) – b2(b – c)(a – b)] + c2(c – a)(c – b)
= a2(a – b)2 + (a2 – b2)(a – b)(b – c) + c2(c – a)(c – b)
= a2(a – b)2 + (a + b)(a – b)2(b – c) + c2(c – a)(c – b) ≥ 0 , since c – a ≥ 0 , c – b ≥ 0.
25.
⎞
⎛ 4 2 ⎞⎛ 4 2 ⎞ ⎛ 4
By CBS inequality, ⎜⎜ ∑ x i ⎟⎟⎜⎜ ∑ y i ⎟⎟ ≥ ⎜⎜ ∑ x i y i ⎟⎟
⎠
⎠ ⎝ i=1
⎠⎝ i=1
⎝ i=1
2
Put x 1 = a , x 2 = b , x 3 = c , x 4 = d
; y1 = a 3 , y 2 = b 3 , y 3 = c 3 , y 4 = d 3
∴
(a + b + c + d) (a3 + b3 + c3 + d3) ≥ (a2 + b2 + c2 + d2)2.
7
26.
Let f(x) = x – ln (1 + x) , then f ' ( x ) = 1 −
1
x
=
> 0 , as x > 0. Therefore f(x) is increasing.
1+ x 1+ x
Since x > 0 , f(x) > f(0) = 0 – ln (1 + 0) = 0 ⇒ x – ln (1 + x) > 0 ⇒ x > ln (1 + x).
(1 + x ) − x = 1 − 1 = 1 + x − x = x > 0
1
x
−
, then g' ( x ) =
Let g(x) = ln(1 + x ) −
1+ x
1+ x
(1 + x )2 1 + x (1 + x )2 (1 + x )2 (1 + x )2
Therefore g(x) is increasing.
Since x > 0 , g(x) > g(0) = ln(1 + 0) −
27.
⇒ ln(1 + x ) >
0
=0
1+ 0
x
.
1+ x
Method 1
(
) (
) (
)
1 − x n 1 − x n +1 (n + 1) 1 − x n − n 1 − x n +1
1 − x n − nx n (1 − x )
−
=
=
n
n +1
n (n + 1)
n (n + 1)
E=
=
(1 − x ) [(1 + x + x 2 + ... + x n −1 ) − nx n ] = (1 − x )2 [1 + 2x + 3x 2 + ... + (n − 1)x n −2 + nx n −1 ]
n (n + 1)
n (n + 1)
> 0.
1 − x n +1 1 − x n
.
<
n +1
n
∴
Method 2
Since 0 ≤ x ≤ 1 ⇒ 0 ≤ xn-1 ≤ 1 and 0 ≤ xn ≤ 1 ⇒ x n = x xn-1 ≤ 1 xn-1 = xn-1 .
Since xn-1 and xn are positive, ∀x, 0 ≤ x ≤ 1
x n ≤ xn-1
∫
⇒
1
1
x n dx ≤ ∫ x n −1dx
x
x
x n +1 1 x n 1
≤
n +1 x n x
⇒
1 − x n +1 1 − x n
≤
n +1
n
⇒
When x = 1 , equality holds.
When x > 1 , x n > xn-1 > 1 ⇒
∫
28.
x
1
x n dx > ∫
x
1
x n −1dx
x n +1 x x n x
>
n +1 1 n 1
⇒
x n +1 − 1 x n − 1
>
⇒
n +1
n
⇒
Let a, m , n > 0 , m > n .
1
1
log 1 + a m < log 1 + a n ⇔ n log 1 + a m < m log 1 + a n ⇔ log 1 + a m
m
n
(
)
(
( ) < (1 + a )
(1 + a ) > 1 ⇔
⇔
(1 + a )
⇔ 1+ am
n
n m
)
(
)
(
)
(
)
n
(
1 − x n +1 1 − x n
.
<
n +1
n
< log 1 + a n
)
m
, since log x is an increasing function.
n
⎡1 + a n ⎤
1+ an
⎢
m ⎥
1
+
a
⎣
⎦
n m
m n
(
)
m−n
>1
(
If a < 1, then
am < an ⇒ 1 + a m < 1 + a n ⇒ 1 + a m
If a = 1, then
(1 + a )
(1 + a )
If a > 1, then
⎡1 + an ⎤
1+ an
⎢
m ⎥
⎣1 + a ⎦
n m
=
m n
n
(
) < (1 + a )
n
n m
⇒
(1 + a )
(1 + a )
n m
m n
>1 .
2m
>1 .
2n
)
m−n
n
⎡ an ⎤
> ⎢ m ⎥ 1+ an
⎣a ⎦
(
)
m−n
(
= a ( n −m ) n 1 + a n
)
m −n
> an
2
− mn
(a )
n m −n
= a0 =1 .
8
29.
⎛1 + y ⎞
1
1
⎟⎟
−
− ln⎜⎜
1− y 1+ y
⎝1− y ⎠
f ( y) =
(a)
f ' ( y) =
=
1
1
1 − y (1 − y ) − (1 + y )(− 1)
1
1
2
+
−
×
=
+
−
2
2
(1 − y) (1 + y) 1 + y
(1 − y)2 (1 + y)2 (1 − y )(1 + y)
(1 − y )2
(1 + y )2 + (1 − y )2 − 2(1 − y 2 ) =
4y 2
2
2
(1 − y ) (1 + y )
(1 − y )2 (1 + y )2
> 0. for 0 < y < 1 .
∴
f(y) is an increasing function.
1
1
⎛1 + 0 ⎞
−
− ln⎜
f(y) > f(0) =
⎟ =0⇒
1− 0 1+ 0
⎝1− 0 ⎠
⎛1 + y ⎞
1
1
⎟⎟ <
.
ln⎜⎜
−
⎝1− y ⎠ 1− y 1+ y
1
1
c
⎛1+ c / x ⎞
−
Put y =
where x > c > 0 , then 0 < y < 1 . By (a) , ln⎜
⎟<
x
⎝1− c / x ⎠ 1− c / x 1+ c / x
(b)
x
x
⎛x +c⎞
−
. For a > b > c, we have
Therefore 0 < ln⎜
⎟<
⎝x −c⎠ x −c x +c
a
a ⎛
a
a
a xdx
a
x
x ⎞
xdx
⎛x +c⎞
∫b ln⎜⎝ x − c ⎟⎠dx < ∫b ⎜⎝ x − c − x + c ⎟⎠dx ⇒ ∫b ln(x + c)dx − ∫b ln(x − c)dx < ∫b x − c − ∫b x + c
30.
a ⎡
a ⎡
a
a
x ⎤
x ⎤
dx < ∫ ⎢ln (x − c ) +
dx ⇒ x ln(x + c ) < x ln (x − c )
⇒ ∫ ⎢ln(x + c ) +
⎥
⎥
b
b
b
b
x + c⎦
x − c⎦
⎣
⎣
⇒ a ln (a + c) – b ln (b + c) < a ln (a – c) – b ln (b – c)
⇒ a ln (a + c) –a ln (a – c) < b ln (b + c) – b ln (b – c)
a
b
a
b
⎛a +c⎞
⎛b+c⎞
⎛a +c⎞
⎛b+c⎞
⎛a+c⎞ ⎛ b+c⎞
<
⇒ a ln⎜
⎟ ⇒ ln⎜
⎟ < ln⎜
⎟ ⇒ ⎜
⎟ < b ln⎜
⎟ ⎜
⎟
⎝ a −c⎠ ⎝ b−c⎠
⎝b−c⎠
⎝a −c⎠
⎝b−c⎠
⎝a −c⎠
Put α = β + γ
⎛ 1 ⎞ ⎡ γ (1) + β(1 + 1 / β ) ⎤
⎛ 1⎞
⎜⎜1 + ⎟⎟ = 1γ ⎜⎜1 + ⎟⎟ < ⎢
⎥
γ+β
⎦
⎝ β⎠ ⎣
⎝ β⎠
β
⎡ γ + β + 1⎤
=⎢
⎥
⎣ γ+β ⎦
∴
β
γ +β
α
1⎞
⎛ α + 1⎞
⎛
=⎜
⎟ = ⎜1 + ⎟
α⎠
⎝ α ⎠
⎝
⎛ 1⎞
Sn = ⎜1 + ⎟
⎝ n⎠
γ +β
, by A.M. > G.M.
α
n
is a monotonic increasing function.
n
1
1 ⎞ ⎛ 1⎞
⎛
For n > 1, Sn = ⎜1 + ⎟ > ⎜1 + ⎟ = 2 . Also,
n ⎠ ⎝ 1⎠
⎝
⎛n⎞ 1 ⎛n⎞ 1
⎛n⎞ 1
n (n − 1)...(n − r + 1) ⎛ 1 ⎞
⎛ 1 ⎞ n (n − 1) ⎛ 1 ⎞
Sn = 1 + ⎜⎜ ⎟⎟ + ⎜⎜ ⎟⎟ 2 + ... + ⎜⎜ ⎟⎟ r + ... = 1 + n ⎜ ⎟ +
⎜ 2 ⎟ + ... +
⎜ r ⎟ + ...
2! ⎝ n ⎠
r!
⎝n⎠
⎝n ⎠
⎝1⎠ n ⎝ 2⎠ n
⎝r⎠n
n×n⎛ 1 ⎞
n × n × ... × n ⎛ 1 ⎞
1 1
1
⎜ 2 ⎟ + ... +
⎜ r ⎟ + ... = 1 + 1 + + + ... + + ...
2! ⎝ n ⎠
r!
2! 3!
r!
⎝n ⎠
< 1 + 1 + 0.5 + 0.16667 + 0.04167 + 0.00833 + 0.00138 +…
= 2.718…
m
⎛ n + 1⎞
m
m
2 m + 4 m + 6 m + ... + (2n ) ≥ n (n + 1) ⇔ 1m + 2 m + 3m + ... + n m ≥ n ⎜
⎟
⎝ 2 ⎠
By Tchebychev’s inequality,
n 1m−1 × 1 + 2 m−1 × 2 + 3m−1 × 3 + ... + n m−1 × n ≥ 1m−1 + 2 m−1 + 3m−1 + ... + n m−1 (1 + 2 + 3 + ... + n )
<1+1+
31.
(
) (
= (1
m −1
)
⎡ n (n + 1) ⎤
+ 2 m−1 + 3m−1 + ... + n m−1 ⎢
⎥
⎣ 2 ⎦
)
….(1)
9
⎡ n (n + 1) ⎤
n 1m − 2 × 1 + 2 m − 2 × 2 + 3m − 2 × 3 + ... + n m − 2 × n ≥ 1m − 2 + 2 m − 2 + 3m − 2 + ... + n m − 2 ⎢
⎥
⎣ 2 ⎦
:
:
:
:
:
:
(
)
n
n
+
1
⎡
⎤
n (1 × 1 + 2 × 2 + 3 × 3 + ... + n × n ) ≥ (1 + 2 + 3 + ... + n )⎢
⎥
⎣ 2 ⎦
Multiply the inequalities,
m −1
m
⎛ n (n + 1) ⎞⎛ n (n + 1) ⎞
m⎛ n +1⎞
n m −1 1m + 2 m + 3m + ... + n m ≥ ⎜
⎟
⎟⎜
⎟ =n ⎜
⎝ 2 ⎠⎝ 2 ⎠
⎝ 2 ⎠
(
) (
(
)
….(2)
…(n)
)
m
32.
⎛ n +1⎞
Therefore 1m + 2 m + 3m + ... + n m ≥ n ⎜
⎟
⎝ 2 ⎠
a 4 + b 4 + c4 a 2a 2 + b 2b 2 + c2c2 a 2 + b 2 + c2 a 2 + b 2 + c2
=
≥
By Tchebychev’s inequality,
3
3
3
3
2
2
⎛ aa + bb + cc ⎞ ⎛ a + b + c a + b + c ⎞ ⎛ a + b + c ⎞
=⎜
⎟
⎟ =⎜
⎟ ≥⎜
3
3
3
3
⎠
⎠ ⎝
⎝
⎠ ⎝
4
4
4
4
∴
27(a + b + c ) > (a + b + c) .
33.
4
Method 1
1
2
13 + 23 + 33 + …. + n3 = n 2 (n + 1)
4
1
2
3
2
2
8(13 + 23 + 33 + …. + n3) – n(n + 1)3 = 8 × n 2 (n + 1) − n (n + 1) = n (n + 1) [2n − n − 1] = n (n + 1) (n − 1)
4
≥ 0, n > 1.
∴
n(n + 1)3 ≤ 8(13 + 23 + 33 + …. + n3) .
Method 2
8k3 – [ k (k + 1)3 – (k – 1) k3 ] = 8k3 – k [k3 + 3k2 + 3k + 1 – k3 + k2 ] = 4k3 – 3k2 – k = k(4k + 1)(k – 1) ≥ 0
∑ [k(k + 1) − (k − 1)k
n
∴
k (k + 1)3 – (k – 1) k3 ≤ 8k3
∴
n(n + 1)3 ≤ 8(13 + 23 + 33 + …. + n3) .
k =1
34.
By A.M. > G.M.,
(i)
(ii)
]≥ 8∑ k
n
3
3
k =1
….
(1)
….
(2)
⎛ 1
1
1 ⎞
+
+ ⎟⎟ ≥ 9
⎝ x1 x 2 x 3 ⎠
x
Put
f(x) = e – (1 + x), then f ’(x) = ex – 1 .
∴ f(x) is increasing.
If x > 0 , f ’(x) > e0 – 1 = 0 .
0
∴ f(x) is decreasing.
If x < 0 , f ’(x) < e – 1 = 0 .
∴ f(x) is a minimum at x = 0 .
∴ f(x) = ex – (1 + x) > f(0) = e0 – (1 + 0) = 0 and ex > 1 + x for x ≠ 0.
9×(1)×(2),
35.
x1 + x 2 + x 3 3
≥ x1x 2 x 3
3
1
1
1
+
+
x1 x 2 x 3
1 1 1
≥3
3
x1 x 2 x 3
3
Put
(x1 + x 2 + x 3 )⎜⎜
⎛
x3 ⎞
⎛ π⎞
g(x) = tan x − ⎜⎜ x + ⎟⎟, x ∈ ⎜ 0, ⎟ , then g’(x) = sec2 x – 1 – x2 = tan2 x – x2 > 0,
3 ⎠
⎝ 2⎠
⎝
⎛ π⎞
since tan x > x, for x ∈ ⎜ 0, ⎟ .
⎝ 2⎠
10
36.
37.
⎛
⎛
x3 ⎞
03 ⎞
∴ g(x) is increasing. Since x > 0, g(x) = tan x − ⎜⎜ x + ⎟⎟ > g(0) = tan 0 − ⎜⎜ 0 + ⎟⎟ = 0.
3⎠
3 ⎠
⎝
⎝
x3
⎛ π⎞
< tan x , x ∈ ⎜ 0, ⎟ .
∴ x+
⎝ 2⎠
3
If b2 < ac , then x2 + 2bx + ac > 0 ∀ x∈
and a ≠ 0.
Put x = 2a,
a(4a + 4b + c) > 0
….
(1)
Put x = a,
a(a + 2b + c)
>0
….
(2)
(1) × (2),
a2(4a + 4b + c) (a + 2b + c) > 0.
Since a2 > 0 and a + 2b + c > 0 , we have 4a + 4b + c > 0.
C nk
1
n!
1
1 n n −1 n − 2
n − k +1 1
2 ⎞ ⎛ k + 1⎞ 1
⎛ 1 ⎞⎛
=
= × ×
×
× ... ×
= × 1 × ⎜1 − ⎟⎜1 − ⎟...⎜1 −
⎟≤
k
k
n
k!(n − k )! n
k! k
k
k
k
k!
n ⎠ k!
⎝ n ⎠⎝ n ⎠ ⎝
n
1
1
1 1
1⎞
⎛
n 1
n 1
n 1
n 1
⎜1 + ⎟ = 1 + C1 + C 2 2 + ... + C k k + ... + C n n ≤ 1 + + + ... + + ... +
n!
k!
1! 2!
n
n
n
n
n⎠
⎝
1
1
1
1
<1+1+
+
+ ... +
+
1× 2 2 × 3
(n − 2)(n − 1) (n − 1)n
1 ⎞ ⎛ 1
1⎞
1
⎛ 1⎞ ⎛1 1⎞
⎛ 1
= 1 + 1 + ⎜1 − ⎟ + ⎜ − ⎟ + ...⎜
−
− ⎟ =1+1+1− < 3
⎟+⎜
n
⎝ 2⎠ ⎝ 2 3⎠
⎝ n − 2 n −1⎠ ⎝ n −1 n ⎠
38.
(i)
(ii)
2(x2 + y2) – (x + y)2 = x2 – 2xy + y2 = (x – y)2 ≥ 0
∴
2(x2 + y2) ≥ (x + y)2 and the equality holds ⇔ (x – y)2 = 0 ⇔ x = y.
2
2
2
2
2
⎡⎛
1⎞ ⎛
1⎞ ⎤
1
1⎞ ⎛ 1 1⎞ ⎛ a +b⎞
⎛
2⎢⎜ a + ⎟ + ⎜ b + ⎟ ⎥ ≥ ⎜ a + + b + ⎟ = ⎜1 + + ⎟ = ⎜1 +
⎟
a⎠ ⎝
b ⎠ ⎥⎦ by ( a ) ⎝
a
b⎠ ⎝ a b⎠ ⎝
ab ⎠
⎢⎣⎝
2
⎛ ⎛ 2 ⎞2 ⎞
⎟ ,
≥ ⎜1 + ⎜
⎜ ⎝ a + b ⎟⎠ ⎟
⎠
⎝
[
]
(i)
(ii)
2
2
1⎞ ⎛
1⎞
25
⎛
.
⎜a + ⎟ + ⎜b + ⎟ ≥
⎝
a⎠ ⎝
b⎠
2
1 x5 − 1
(1)
If x ≥ 1, x 2 − 3 =
≥0 ,
since x5 ≥ 1.
3
x
x
1
x5 − 1
≤0,
(2)
If 0 < x < 1, x 2 − 3 =
since x5 ≤ 1.
x
x3
1
1
x −1
1 ⎞
⎛
Combining (1) and (2), (x − 1)⎜ x 2 − 3 ⎟ ≥ 0 ⇒ x 2 (x − 1) ≥ 3 ⇒ x 3 − x 2 ≥ 2 − 3
x
x
x
x
⎝
⎠
The equality holds when x = 1.
2
= 1 + 2 2 = 25
39.
2
⎛a +b⎞
sin ce ⎜
⎟ ≥ ab .
⎝ 2 ⎠
∴
Let a, ar, ar2 be the sides of the triangle.
Since the sum of any two sides > the third side.
….
(1)
ar + ar2 > a
….
(2)
ar + a > ar2
From (1),
Since r > 0,
From (2),
Since a > 0, r + r2 > 1, r2 + r – 1 > 0
∴
r>
∴ r>
5 −1
− 5 −1
or r <
….(3)
2
2
5 −1
2
Since a > 0, r2 – r – 1 < 0.
∴ r<
1+ 5
5 −1
or r >
…. (4)
2
2
11
Combining (3) and (4),
40.
41.
5 −1
2
2
x < 1 or x > 2.
B
C
1 − tan tan
π − (B + C )
A
B+C
⎛π B+C⎞
2
2
= tan⎜ −
tan = tan
=
⎟ = cot
B
C
2
2
2 ⎠
2
⎝2
tan + tan
2
2
A⎛
B
C⎞
B
C
tan ⎜ tan + tan ⎟ = 1 − tan tan
⇒
2⎝
2
2⎠
2
2
A
B
B
C
C
A
⇒
tan tan + tan tan + tan tan = 1
2
2
2
2
2
2
2
Now,
⇒
2
….
(1)
2
A
B⎞ ⎛
B
C⎞ ⎛
C
A⎞
⎛
⎜ tan − tan ⎟ + ⎜ tan − tan ⎟ + ⎜ tan − tan ⎟ ≥ 0
2
2
2
2
2
2⎠
⎠ ⎝
⎝
⎠ ⎝
A
B
C⎞ ⎛
A
B
B
C
C
A⎞
⎛
2⎜ tan 2 + tan 2 + tan 2 ⎟ − 2⎜ tan tan + tan tan + tan tan ⎟ ≥ 0
2
2
2⎠ ⎝
2
2
2
2
2
2⎠
⎝
A
B
C⎞
⎛
2⎜ tan 2 + tan 2 + tan 2 ⎟ − 2(1) ≥ 0
2
2
2⎠
⎝
A
B
C
tan 2 + tan 2 + tan 2 ≥ 1
2
2
2
⇒
⇒
42.
5 +1
<r<
⎛a +b⎞
Since ⎜
⎟
⎝ 2 ⎠
a +b
>
( ab )
a +b
= (ab )
a +b
2
a +b
It remains to show (ab ) 2 > a b b a
….
Since (1) is symmetric in a, b we can let a ≥ b > 0 .
a +b
(ab) 2
a −b
⎛a⎞ 2
= ⎜ ⎟ >1,
b a
a b
⎝b⎠
Result follows.
since
a
≥ 1,
b
(1)
a−b
≥0 .
2
12