Page 95 - Elementary Algebra Exercise Book I

ELEMENTARY ALGEBRA EXERCISE BOOK I inequAlities

√ √
3.18 If a, b, c > 0, show 2( a+b − ab) ≤ 3( a+b+c − 3 abc).
2 3
√ √ √ √ √
√
√
√
3
3
Proof: 2( a+b √ − ab) ≤ 3( a+b+c √ − abc) ⇔ a+b−2 ab ≤ a+b+c−3 abc ⇔ c+2 ab ≥
3
a+b+c
3
a+b
2( √ 2 − ab) ≤ 3( 3 − abc) ⇔ a+b−2 ab ≤ a+b+c−3 abc ⇔ c+2 ab ≥
3 2
√ 3 abc 3 √ √ √ √ √ √
3
3
3
3 abc . We only need to show the last inequality. c +2 ab = c + ab + ab ≥ 3 c · ab · ab =3 abc
√ √ √ √ √ √
3
c +2 ab = c + ab + ab ≥ 3 3 c · ab · ab =3 abc .
x+3
2
3.19 Given the function f(x) = 4 +8 , (1) find the maximum value of f(x), (2) show
x
2
f(a) <b − 4b + 11 for any real numbers a, b .
2
√ √
x+3
8
2
Solution: (1) f(x) = 4 +8 = x 8 8 ≤ √ 8 8 = √ = 2, thus f(x) max = 2.
x
x
2 + x
2 2 2 · x 4 2
2
√ √
11
2
2
2
(2) Since f(a) ≤ 2 and b − 4b + 11 =(b − 2) + 3 ≥ 3 2 > 2, we have f(a) <b − 4b +
2
2
2
for any real numbers a, b .
√ √
3.20 Show 2( n +1 − 1) < 1+ √ + √ + ··· + √ < 2 n for any n ∈ N .
1
1
1
2 3 n
√ √
1
2
1
2
1
Proof: √ = 2 k k+ k+1 = 2( k +1 − k) . Let m = 1+ √ + √ + ··· + √ 1 n ,
√
√ > √
3
k
2
√ √ √ √ √ √
then m> 2( 2 − 1+ 3 − 2+ ··· + n +1 − n) = 2( n +1 − 1),
√ √ √ √ √ √
and m< 2(1 − 0+ 2 − 1+ 3 − 2+ ··· + n − n − 1) = 2 n .
√ √
1
1
1
Hence, 2( n +1 − 1) < 1+ √ + √ + ··· + √ < 2 n
2 3 n
2
2
2
1
3.21 Given a + b + c =1, show − ≤ ab + bc + ca ≤ 1.
2
2
2
2
2
2
2
2
2
2
Proof: a + b ≥ 2ab, b + c ≥ 2bc, c + a ≥ 2ca , add them up to obtain ab + bc + ca ≤ a + b + c =1

2
2
2
ab + bc + ca ≤ a + b + c =1 . Since (a + b + c) ≥ 0 , a + b + c + 2(ab + bc + ca) ≥ 0 , then
2
2
2
2
1
ab + bc + ca ≥− (a + b + c )= − , thus − ≤ ab + bc + ca ≤ 1.
1
1
2
2
2
2 2 2
c
b
3
a
3.22 Given a, b, c > 0, show a+b + b+c + c+a ≥ .
2
1
c
1
b
1
a+b+c
a
=
Proof: a+b + a b+c + b c+a a+b+c + a+b+c + a+b+c − 3 = (a + b + c)( 1 a+b + 1 b+c + 1 c+a ) − 3=
a+b+c
b+c
a+b+c
c+a
a+b
c
+
) − 3=
+
+
+
+ a+b+c
+ a+b+c
1
a
− 3 = (a + b + c)( 1
b
c
1
= a+b+c
3 +
1 a+b + b+c + c+a = a+b + b+c 1 + c+a − 1 3 = (a + b + c)( a+b b+c + c+a) − 3=
1
1
[(a + b)+(b + c) +(c + a)](
+
b+c
c+a
· 3 a+b
1 [(a a+b b+c c+a a+b b+c a+b + c+a + 1 c+a ) − 3 ≥ 1 2 (a + b)(b + c)(c + a) ·
b+c
1
2
1
+
3
1 + b)+(b + c) +(c + a)]( 1 b+c + c+a) − 3 ≥ 2 · 3 (a + b)(b + c)(c + a) ·
1
1
) − 3 ≥ 1
· 3 3
a+b +
2[(a + b)+(b + c) +(c + a)](
(a + b)(b + c)(c + a) ·
a+b
2 3 1 1 1 − 3= 9 − 3= 3 b+c c+a 2
3 · ·
3 1 a+b 1 b+c 1 c+a 9 2 3 2
1
− 3= 3
1
1
− 3= 9
3
3 3 a+b · · b+c · · c+a − 3= 2 − 3= .
2
a+b b+c c+a 2 2
√
3.23 Solve the inequality 2x +5 >x +1.
5
Solution: To make the square root valid, we need 2x +5 ≥ 0 ⇔ x ≥− . When x +1 < 0,
2
√
i.e. x< −1, we have 2x +5 ≥ 0 >x +1, thus the original inequality has the solution
− ≤ x< −1. When x ≥−1, the original inequality has the solution −1 ≤ x< 2. The union
5
2
5
of these two solution sets provides the solution of the original inequality: {x|− ≤ x< 2}.
2
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