Solutions to Algebraic Inequalities
1. Which is larger? 3111 or 1714
3111 < 3211 = 255 < 256 = 1614 < 1714
Hence 1714 is larger.
2. (n3)100 > (35)100
n3 > 35 = 243
Now, 33 = 27, 43 = 64, 53 = 125, 63 = 216 < 243, 73 = 343 > 243
Hence the required smallest value is n = 7.
3. For all x λ ,
2
10 x + 1 10 x +1 + 1
> x+2
Û 10 x + 1 10 x +2 + 1 > 10 x +1 + 1
x +1
10 + 1 10 + 1
(
(10 + 1) (10
x
x+2
)(
) (
)
) (
2
+ 1 - 10 x +1 + 1 > 0
(
)
)
102 x + 2 + 10 x + 2 + 10 x + 1- > 0 102 x + 2 + 2 ´ 10 x +1 + 1 > 0
10
x+2
(
- 2 ´ 10
x +1
+ 10 > 0
x
)
10 10 - 2 ´ 10 + 1 > 0
x
2
81 ´ 10 x > 0
4. Since both numbers are positive, it suffices to show that
( ) ( )
8
72
8!
<
9
72
9!
(8!) < (9!)
9
8
Since 9! = 9  8!, we can transform the inequality to (8!)9 < 98  (8!)8.
Dividing both sides of the inequality by (8!)8, we have
8! < 98
This last inequality holds since it is the product of 8 inequalities:
1 < 9, 2 < 9, …, 8 < 9
Homework Problems
1. Ans: 105
3p + 7q + r = 315
4p + 10q + r = 420
(1)  3:
(2)  2:
(1)
(2)
9p + 21q + 3r = 945
8p + 20q + 2r = 840
p + q + r = 105
2. Four numbers a, b, c, d satisfy these two conditions:
a+b+c+d=0
and
abc + bcd + cda + dab = 91
What is the value of a3 + b3 + c3 + d3?
a+b+c+d=0
(1)
From (1), a + b = –(c + d)
Taking cubes on both sides, we have
(a + b)3 = –(c + d)3
a3 + b3 + 3ab(a + b) = –c3 – d3 – 3cd(c + d)
a3 + b3 + c3 + d3 = –3ab(a + b) – 3cd(c + d)
= 3ab(c + d) + 3cd(a + b)
= 3  91
= 273
3. Ans: 4
p+q=1
q=1–p
1
1 q+ p
+ =
p q
pq
1
=
pq
1
=
p(1 - p)
=
1
-( p - p)
=
1
2
é
1 1ù
- ê p2 - p + - ú
4 4û
ë
1
=2
éæ
1ö
1ù
êç p - ÷ - ú
2ø
4ú
êëè
û
1
=
2
1 æ
1ö
- p- ÷
4 çè
2ø
2
1 æ
1ö
1
-çp - ÷ £
Now,
4 è
2ø
4
So
1 1 1
+ ³ =4
1
p q
4
Hence the least value of
(
)(
1 1
+ is 4.
p q
)
5. (a) 2x + y x - 2y = 7 = 7  1
Þ
2x + y = 7
x – 2y = 1
(
)(
)
(b) x - y 3x - 2y = 13
x=3
y=1
= 1  13 or –13  –1
x- y =1
3x - 2 y = 13
Þ
x = 11
y = 10
Þ
x = 25
y = 38
x - y = -13
3x - 2 y = -1
6. 3 solutions.
((x – 2011)2 – 2010)2 = 20102
(x – 2011)2 – 2010 = 2010
(x – 2011)2 = 2010  2010
(x – 2011)2 = 4020
or
(x – 2011)2 = 0
(x – 2011) = ± 4020
x = 2011 
x = 2011
4020
Thus there are 3 different solutions.
7. Ans: 12
Since 6(pq + qr + rp) = 2(p + q + r)2 – (p – q)2 – (p – r)2 – (q – r)2, the largest value of
pq + qr + rp is obtained when p = q = r = 2.
Therefore the answer is 12.
10. Ans: 4
r2 = pq – 16
= (8 – q)q – 16
= –(q2 – 8q + 16)
= –(q – 4)2
r2 + (q – 4)2 = 0
 r = 0 and q = 4
q = 4, p = 8 – 4 = 4
p+r=4+0=4
12. Ans: 10
x5 – 101x3 – 999x2 + 100 900 = 0
(x2 – 101)(x3 – 999) + 1 = 0
Clearly, the only integral solution is x = 10.