SEBA Class VIII Advanced Mathematics CHAPTER 2: INEQUALITY AND INEQUATIONS

 CHAPTER 2: INEQUALITY AND INEQUATIONS

EXERCISE 2(a)

Prove that:

1. (i)

Solution:

We know that,

(ii) If a > 0, b > 0, a > b then

Solution:

Given,

a > 0, b > 0

Now, 

a > b

2. (i)

Solution:

We know that,

…………. (1)

…………. (2)

(1)×(2)⇒

(ii)

Solution:

We know that,

……………… (1)

and

……………… (2)

(1) × (2)⇒

3. a2 + b2 ≥2ab

Solution:

We know that,

(a – b)2 ≥ 0

⇒ a2 – 2ab + b2 ≥ 0

⇒ a2 + b2 ≥ 2ab

4. If a > 0, b> 0, x > 0 and a < b, then

Solution:

Given,

a < b

⇒ ax < bx [Multiplying both sides by x]

⇒ ab + ax < ab + bx [Adding ab to both sides]

⇒ a(b + x) < b(a + x)

5.

Solution:

We know that,

a2 + b2 ≥ 2ab

⇒ a2 + b2 + a2 + b2 ≥ a2 + b2 + 2ab [Adding a2 + b2 on both sides]

⇒ 2(a2 + b2) ≥ (a + b)2

………………. (1)

Similarly,

………………. (1)

and

………………. (1)

(1) + (2) + (3) ⇒

6. (i) If a > 1, b > 1 then (a + 1)(b + 1) < 2(ab + 1)

Solution:

⸪ a > 1

⇒ a – 1 > 0

⸪ b > 1

⇒ b – 1 > 0

⸫(a – 1)(b – 1) > 0

⇒ a(b – 1) – (b – 1) > 0

⇒ ab – a – b + 1 > 0

⇒ ab + 1 > a + b

⇒ ab + 1 + ab + 1 > ab + 1 + a + b [Adding ab + 1 to both sides]

⇒ 2ab + 2 > ab + a + b + 1

⇒ 2(ab + 1) > a(b + 1) + (b + 1)

⇒ 2(ab + 1) > (b + 1)(a + 1)

⇒ 2(ab + 1) > (a + 1)(b + 1)

⇒ (a + 1)(b + 1) < 2(ab + 1)

(ii) If a, b, c ∈ R. Prove that (a + b)(b + c)(c + a) ≥ 8abc

Solution:

We know that,

…………….. (1)

Similarly,

…………….. (2)

and

…………….. (3)

Multiplying (1), (2) and (3), we get,

7. (i) a3 + b3 ≥ a2b + ab2

Solution:

We know that,

……………. (1)

and

……………. (2)

Dividing (1) by (2) we get,

⇒ a3 + b3 ≥ ab(a + b)

⇒ a3 + b3 ≥ a2b + ab2

(ii) a4 + b4 ≥ a3b + ab3.  

Solution:

We know that,

……………. (1)

and

 

……………. (2)

Dividing (1) by (2) we get,

⇒ a4 + b4 ≥ ab(a2 + b2)

⇒ a4 + b4 ≥ a3b + ab3

8. If a > b > 0 and c > 0, then prove that:

(i)

Solution:

Given,

a > b

⇒ ac > bc [Multiplying both sides by c]

⇒ ab + ac > ab + bc [Adding ab to both sides]

⇒ a(b + c) > b(a + c)

(ii) a3 + b3 > a2b + ab2

Solution:

We know that,

……………. (1)

and

……………. (2)

Dividing (1) by (2) we get,

⇒ a3 + b3 ≥ ab(a + b)

⇒ a3 + b3 ≥ a2b + ab2

(iii) a3 – b3 ≥ a2b – ab2

Solution:

Given,

a > b

⇒ a – b > 0

Now,

a2 + b2 ≥ 0

⇒ a2 + b2 + ab ≥ ab [Adding ab to both sides]

⇒ (a – b)(a2 + b2 + ab) ≥ ab(a – b) [Multiplying both sides by a – b]

⇒ a3 – b3 ≥ a2b – ab2

(iv)

Solution:

Given

a > b

⇒ a2 > b2

⇒ 2a2 > 2b2 [Multiplying both sides by 2]

⇒ ab + 2a2 > ab + 2b2 [Adding ab to both sides]

⇒ a(b + 2a) > b(a + 2b)

9. If a, b, c, d are positive and then prove that

Solution:

Given,

[Adding to both sides]

[Multiplying both sides by d]

……………. (1)

Also,

[Adding to both sides]

[Multiplying both sides by b]

……………. (2)

From (1) and (2) we get,

10. If a, b, c are positive then show that

(i)

Solution:

⸪a, b, c > 0,

⸫ ab, bc, ca > 0

We know that,

……………….. (1)

Similarly,

……………….. (2)

and

……………….. (3)

Adding (1), (2) and (3) we get,

(ii)

Solution:

⸪a, b, c > 0

We know that,

(iii)

Solution:

Given,

a, b, c > 0

⸫ b2 + c2 ≥ 0

⇒ b2 + 2bc + c2 ≥  2bc [Adding 2bc to both sides]

⇒ (b + c)2 ≥  2bc

………… (1)

Similarly,

………… (2)

and

………… (3)

Adding (1), (2) and (3) we get,

11. If a > 0, a ≠1, then show that

(i)

Solution:

We know that

(a – 1)2 > 0

⇒ (a – 1)2(a2 + a + 1) > 0 [Multiplying both sides by a2 + a + 1]

⇒ (a – 1)(a – 1)(a2 + a + 1) > 0

⇒ (a – 1)(a3 – 1) > 0

⇒ a(a3 – 1) – 1(a3 – 1) > 0

⇒ a4 – a – a3 + 1 > 0

⇒ a4 + 1 > a3 + a

⇒ a4 + 1 > a(a2 + 1)

[Dividing both sides by a]

(ii)

Solution:

Given,

a > 0

⸫a4, a3, a2, a > 0