Question

Suppose a1, a2. . . , an are n positive real numbers with a1a2 . . . an = 1.

Then the minimum value of (1 + a1)(1 + a2) . . . (1 + an) is

(A) 2n,

(B) 22n

(C) 1,

(D)None of the above.

Answer 1

∵ a₁ , a₂ , a₃ , ....... an are n positive real numbers
∴ we have to use AM ≥ GM
Where AM is Arithmetic mean and GM is Geometric mean.

Now, AM of a₁ , a₂, a₃, a₄, ........., an = (a₁ + a₂ + a₃ + ..... + an)/n
GM of a₁ , a₂, a₃, ........, an = (a₁.a₂.a₃.a₄....an)^(1/n)
∴ (a₁ + a₂ + a₃ + ..... + an)/n ≥ (a₁.a₂.a₃.a₄....an)^(1/n)
∵ a₁.a₂.a₃.a₄......an = 1
So, (a₁ + a₂ + a₃ + ..... + an)/n ≥ 1
⇒a₁ + a₂ + a₃+ ...... + an ≥ n
minimum value of a₁ + a₂ + a₃+ ...... + an = n

Now, take , (1 +a₁), (1 + a₂), (1 + a₃) ....... (1 + an)
AM = {(1 + a₁) + (1 + a₂) + (1 + a₃) + ..... + (1 + an)}/n
GM = {(1 + a₁)(1 + a₂)(1 + a₃)(1 + a₄)......(1 + an)}^{1/n}

Now, AM ≥ GM
{(1 + a₁) + (1 + a₂) + (1 + a₃) + ..... + (1 + an)}/n ≥ {(1 + a₁)(1 + a₂)(1 + a₃)(1 + a₄)......(1 + an)}^{1/n}
⇒{(1 + 1 + 1 + 1 ....+ n times) + (a₁ + a₂ + a₃ + .....+ an)}/n ≥ {(1 + a₁)(1 + a₂)(1 + a₃)(1 + a₄)......(1 + an)}^{1/n}
⇒(n + n)/n ≥{ (1 + a₁)(1 + a₂)(1 + a₃)(1 + a₄)......(1 + an)}^{1/n}
⇒2 ≥ {(1 + a₁)(1 + a₂)(1 + a₃)(1 + a₄)......(1 + an)}^{1/n}
∴ (1 + a₁)(1 + a₂)(1 + a₃)(1 + a₄)......(1 + an) ≤ 2ⁿ

Hence, minimum value of (1 + a₁)(1 + a₂)(1 + a₃)(1 + a₄)......(1 + an) = 2ⁿ
Option (A).is correct

Answer 2

Step-by-step explanation:

GIven : a₁,a₂,.....an are in continued propotion

To find : show that a₁/aₙ = (a1/a₂)ⁿ⁻¹  ( correction in Question)

Solution:

a₁  , a₂  , a₃  , a₄ ......................aₙ  are in continued proportion

if a₂ = ka₁    => a₁/a₂  = 1/k

=> aₙ  = k aₙ₋₁

LHS =  a₁/aₙ

aₙ  = k aₙ₋₁

aₙ₋₁ = kaₙ₋₂

=> aₙ = k * kaₙ₋₂

=> aₙ = k²aₙ₋₂

=> aₙ = k³aₙ₋₃

=> aₙ = kⁿ⁻¹aₙ₋₍ₙ₋₁₎ =  kⁿ⁻¹a₁

=> LHS =  a₁/ kⁿ⁻¹a₁

= 1/ kⁿ⁻¹

= (1/k)ⁿ⁻¹

a₁/a₂  = 1/k

= (a1/a₂)ⁿ⁻¹

= RHS