Page 105 - Elementary Algebra Exercise Book I

ELEMENTARY ALGEBRA EXERCISE BOOK I inequAlities

1 2 1 2 25
3.53 Let a> 0,b > 0 and a + b =1, show (a + ) +(b + ) ≥ .
a b 2
√ √
1 1 1 ≥ 4.
Proof 1: 1= a + b ≥ 2 ab ⇒ ab ≤ ⇒ ab ≤ ⇒
2 4 ab
And (a+ ) +(b+ ) ) ≥ [ ≥ [ a+ +b+ ] = (1 ++ 1 1 + ) = (1 ++ 1 2 2 2525,
1 1
1 2 1 2
1 2 1 2
1 1
a+ +b+
(a+ ) +(b+
b 2 b 2
1 1
1 2 1 2
1
1 1
+ ) = (1
)
]
a a
b b
a a
) ≥≥
2 2 2 2 4 = (1 a a b b 4 4 abab 4 4
4
1 2 1 2 25
thus (a + ) +(b + ) ≥ .
a b 2
2
1
2
2
2
2
1 2
2
2
2
2
1 2
2
Proof 2: Let a = sin α, b = cos α , then (a+ ) +(b+ ) = (sin α+csc α) +(cos α+sec α) ≥ (sin α+csc α+cos α+
2
a
b
2
2
1
2
2
2
1
2
4
1
1
1
1
sec α) = (1+ sin α cos α 2 1 2 sin α cos α ) = (1+ sin 2α ) = (1+4 csc 2α) ≥
2 +
2 ) = (1+
2
2
2
2
2
2
2
1 11
2
2 2
2 2
2 2
2 2
22
1 2 1 2
2 2
2
2 2
2 25 2
1 2 1 2
(a+ ) +(b+ ) = (sin α+csc α) +(cos α+sec α) ≥ (sin α+csc α+cos α+
(1 + 4) =
(a+ ) +(b+ ) = (sin α+csc α) +(cos α+sec α) ≥ (sin α+csc α+cos α+
2
2
2 2
2
2
1 2
2 2 2 1
1
2
1 2
2
2 2
2 2
2
2
2
b1 2
2 ) +(b+ ) = (sin α+csc
2
2
2
a 1 2
(a+
2
b
2 2 (a+ ) +(b+ ) = (sin α+csc α) +(cos α+sec α) ≥ (sin α+csc α+cos α+
a
1 1
4 4
2 2
1
2 2
2 2
1
1
1 12
2 2
1 2
1 1 2
1
1 1 2
2
2
2
2
2
2
) = (1+ +
2 ) =
a
sec α) =
(a+
) = (1
sec α) = (1+ (1+ b 1 b b 1 2 1 cos α α 1 2 2 2 1 1 α) +(cos α+sec α) ≥ (sin α+csc α+cos α+
) = (1+4 csc 2α) ≥α+
1 a ) +(b+ ) = (sin α+csc α) +(cos α+sec α) ≥ (sin α+csc α+cos
2
2 + +
2 ) = (1+ (1+
2
2
) = (1+4 csc 2α) ≥
2
2 2
1
2 2
2
2 2
2 2 1
1
sin α cos α α 2
cos 1
2
sin 4
1
2
2 21
1
2
2 1
sin 2α 2α4 2
2 sin α cos 1
2 a
2
1
sin α sin α
2 2 1
2
2
sec α) =
1 sec α) = (1+ 2 +
2 + 2 ) = (1+
2 (1+
) = (1+4 csc 2α) ≥
25
2
1
2
1
2
) = (1+ 2 4
2
1
2 2
2
1
1
(1 + 4) =
1
) =
2 +
sec α)
sin α cos α ) = (1+
2
1 (1 + 4) == (1+ sin α cos α 2 ) = (1+ 2 2 1 2 2 α ) = (1+ sin 2α ) = (1+4 csc 2α) ≥
cos α ) = (1+
2
sin α
2
2
sin 2α 2
sin α cos
25 2
2
2
2 (1+4 csc 2α) ≥
25
2
2
2
2
2 1
2
2
2 1 (1 + 4) = 2 2 2 25 sin α cos α 2 sin α cos α 2 sin 2α 2
2 .
2
2 1 (1 + 4) = 25
2
2 (1 + 4) =
2 2
3.54 Let a,b,c,d,m,n be positive real numbers,
√ √ √
d
P = ab + cd, Q = ma + nc b + . Compare P and Q .
m n
√
d
2
2
Solution: P = ab + cd +2 abcd, Q =(ma + nc)( b + )= ab + cd + nbc + mad . Since
m n m n
√
2
2
nbc + mad ≥ 2 nbc · mad =2 abcd , then P ≤ Q . Because P, Q are positive, we have P ≤ Q .
m n m n
3.55 Show the inequality (a + b) ≤ 128(a + b ).
8
8
8
Proof: (a − b) ≥ 0 ⇒ a + b ≥ 2ab. Similarly we have a + b ≥ 2a b ,a + b ≥ 2a b .
2
8
2 2
8
4
4
2
4 4
2
Add a + b ,a + b ,a + b to the above three inequalities respectively to obtain
8
2
4
2
8
4
2(a + b ) ≥ (a + b) , 2(a + b ) ≥ (a + b ) , 2(a + b ) ≥ (a + b ) ) . The last inequality leads
22
4 4
22
8 8
22
2
2 2 2
8 8
2 2
4
4 2 2
4 4
44
2(a + b ) ≥ (a + b) , 2(a + b ) ≥ (a + b ) , 2(a + b ) ≥ (a + b

2 2
2 2
8 8 8 8 4 4 4 2 4 2 4 4 4 4 2 2 22 2 2 2 2 2 2 2 2 2 2

128(a + b ) ≥ 64(a + b ) = 16[2(a + b )] ≥ 16[(a + b ) ] = {[2(a + b )] } ≥
to 128(a + b ) ≥ 64(a + b ) = 16[2(a + b )] ≥ 16[(a + b ) ] = {[2(a + b )] } ≥
2
2 2
2
4
2 2 2
4
2
4
4 2
8
8
2
128(a + b
2 2 2 2 2 2 ) ≥ 64(a + b ) = 16[2(a + b )] ≥ 16[(a + b ) ] = {[2(a + b )] } ≥
{[(a + b) ] } =(a
8
{[(a + b) ] } =(a + b) + b) 8
{[(a + b) ] } =(a + b) .
2 2 2
8
2
3.56 Given the function f(x − 3) = log x 2 2 (a> 0,a =1 ) that satisfies
a 6−x
f(x) ≥ log 2x . Find the domain of the function f(x) .
a
3+t
Solution: Let x − 3= t , then x = 3+ t . Substitute it into the function: f(t) = log a 3−t , thus
2
2
f(x) = log 3+x . Then the inequality f(x) ≥ log 2x is equivalent to log 3+x ≥ log 2x .
a 3−x a a 3−x a
⎧
3+x > 0
⎨ 3−x
If a> 1, then 3+x ≥ 2x ⇒ x ∈ (0, 1) ∪ [− , 3).
3
3−x 2
⎩
x> 0
⎧
3+x >
< 0
⎨ 3−x
If 0 <a < 1, then 3+x ≤ 2x ⇒ x ∈ [1, ) .
3
3−x 2
⎩
x> 0
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