Question

If a, b, c and d are four positive real numbers such that abcd = 1, what is the minimum value of
(1 + a)(1 + b)(1 + c)(1 + d)?
(1) 4 (2) 1 (3) 16 (4) 18

Answer 1

Answer is 18
Hope it will help you.

Answer 2

Answer:

16

Step-by-step explanation:

(1+a)(1+b)(1+c)(1+d)

= (1 + ab + a + b)(1 + cd + c + d)

= 1 + cd + c + d + ab + abcd + abc + abd + a + acd + ac + ad + b + bcd + bc + bd

= 1 + abcd + a + b + c + d + ab + ac + bc + bd + ad + cd  + abc + abd + acd + bcd

abcd = 1

so abc + d ≥2 ,  abd + c  ≥2 , bcd + a  ≥2  , acd + b ≥2

≥  1 + 1  + 2 + 2 + 2 + 2 + ab + ac + bc + bd + ad + cd

abcd = 1  

so ab + cd ≥ 2 , ac + bd ≥ 2 , ad + bc  ≥ 2

≥  10 + 2 +2 + 2

≥  16

So 16 is the right answer

Verification let say a , b , c , d all are = 1

abcd = 1

(1+a)(1+b)(1+c)(1+d) = 2 * 2* 2* 2 = 16