BITSAT 2005 Solved Question Paper

Given that, it is given
αβγδ=1 ......(i)
As, we know A.M. ≥ G.M.
⇒

1+α

2

≥√α
⇒1+α≥√α ......(ii)
Similarly, 1+β≥2√β .......(iii)
1+γ≥2√γ ........(iv)
and 1+δ≥2√δ ........(v)
Multiplying Eqs. (ii), (iii), (iv) and (v), we get
(1+α)(1+β)(1+γ)(1+δ) ≥16√αβγδ
⇒(1+α)(1+β)(1+γ)(1+δ)=16