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Question

# Let α,β,γ and δ are four positive real number such that their product is unity, then the least value of (1+α)(1+β)(1+γ)(1+δ) is:

A
6
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B
16
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C
0
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D
32
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Solution

## The correct option is B 16It is given that the product of α,β,γ and δ is unity that is αβγδ=1.Now consider (1+α)(1+β)(1+γ)(1+δ) as follows:(1+α)(1+β)(1+γ)(1+δ)=1+α+β+γ+δ+αβ+αγ+αδ+βγ+βδ+γδ+αβγ+βγδ+αγδ+αβδ+αβγδ=1+(α+β+γ+δ)+(αβ+αγ+αδ+βγ+βδ+γδ)+(αβγ+βγδ+αγδ+αβδ)+αβγδ=1+(α+β+γ+δ)+(αβ+αγ+αδ+1αδ+1αγ+1αβ)+(1δ+1α+1β+1γ)+1 ....... {∵αβγδ=1} =2+(α+1α)+(β+1β)+(γ+1γ)+(δ+1δ)+(αβ+1αβ)+(αγ+1αγ)+(αδ+1αδ)≥2+2+2+2+2+2+2+2{∵x+1x≥2forx≥0}≥16Hence, the least value of (1+α)(1+β)(1+γ)(1+δ) is 16.

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