Page 100 - Elementary Algebra Exercise Book I

ELEMENTARY ALGEBRA EXERCISE BOOK I inequAlities

1
1
1
3.39 If a, b, c are distinct positive numbers, show + + 1 c > √ + √ 1 ca + √ 1 ab .
a
b
bc
√ √ √ √
√
√
√
√ √ √ √ 2(bc+ca+ab)−2(a bc+b ca+c ab)
√
bc+ca+ab−(a bc+b ca+c ab)
1
1
1
1
1
1 bc+ca+ab−(a bc+b ca+c ab)
1
1
1
Proof 1: + + −( √ + √ + √ )= = 2(bc+ca+ab)−2(a bc+b ca+c ab) = =
1
1
1
=
+
a + −( √ + √ + √ )=
2abc
a √ b c √ b c bc √ bc ca ca ab ab abc abc 2abc
√
√
√
√
√ √ ( ab− bc) +( bc− ca) +( ca− ab) 2
√
2
√
2 √
2
2
( ab− bc) +( bc− ca) +( ca− ab) 2 > 0
abc > 0.
abc
√
√ a √ bc+b ca+c √
√
√
√
√ √
1 1
1
a bc+b ca+c ab ab
1
1 1
1
1
bc+ca+ab
⇔ bc + ca + ab > >
1
1 +
⇔ bc + ca + ab
bc+ca+ab
a bc+b ca+c ab
1
⇔ ⇔
+ √ 1 + √
> √ 1 > √
Proof 2: + + 1 + 1 + √ 1 + √ ca 1 ⇔ ab bc+ca+ab > > > ⇔ bc + ca + ab >
+ √ 1 + √
+ +
> √ c
b
a
a a √ b b c c √ bc √ bc ca ca ab ab abc abc abc abc abc
√ bc
√
√ √ √ √ √ √ √ √ √ abc √ √ √ √ √ √ √ √ √ √ 2

2
bc) +
a bc + b ca + c ab ⇔ 2(bc + ca + ab) > 2(a bc + b ca + c ab) ⇔ ( ab −
a bc + b ca + c ab ⇔ 2(bc + ca + ab) > 2(a bc + b ca + c ab) ⇔ ( ab − bc) + 2 bc) +
√
√
a bc + b ca + c ab ⇔ 2(bc + ca + ab) > 2(a bc + b ca + c ab) ⇔ ( ab −
√
√ √ √ √ √ √ 2 √ √ √ 2
ca) +( ca − ab) ≥ 0 which is obviously valid.
ab) ≥ 0
( bc ( bc − 2 2 ca) +( ca − 2 2 ab) ≥ 0
( bc − −
ca) +( ca −
√ √ √ √ √ √
2
2
2
Proof 3: Since a, b, c are distinct positive numbers, then ( ab − bc) > 0, ( bc − ca) > 0, ( ca − ab) > 0

√ √ √ √ √ √ √ √ √ √ √ √ √ √ √ √ √ √
2 2 2
( ab − bc) > 0, ( bc − ca) > 0, ( ca − ab) > 0. Add them up to obtain 2(ab+bc+ca)−2(a bc+b ca+c ab) > 0 ⇒ ab+bc+ca > a bc+b ca+c ab ⇒
ab+bc+ca > a bc+b ca+c ab ⇒
2(ab+bc+ca)−2(a bc+b ca+c ab) > 0 ⇒
√ √ √ √ √ √ 1 1 1 1 1 1 1 1 1 1 1 1
2(ab+bc+ca)−2(a bc+b ca+c ab) > 0 ⇒ ab+bc+ca > a bc+b ca+c ab ⇒ + + c > √ + √ ca + √ ab .
+ + > √ + √
+ √
bc
b c
a
ca
a b
bc
ab
1
1 + + 1 > √ + √ 1 + √ 1
1
a b c bc ca ab
3.40 Real numbers x,y,z satisfy the inequalities |x| ≥|y + z|, |y|≥ |z + x|, |z| ≥|x + y| .
Show x + y + z =0.
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