Question 31

# 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)?

Solution

Since the product is constant,

(a+b+c+d)/4 >= $$(abcd)^{1/4}$$

We know that abcd = 1.

Therefore, a+b+c+d >= 4

$$(a+1)(b+1)(c+1)(d+1)$$
= $$1+a+b+c+d+ ab + ac + ad + bc + bd + cd+ abc+ bcd+ cda+ dab+abcd$$
We know that $$abcd = 1$$
Therefore, $$a = 1/bcd, b = 1/acd, c = 1/bda$$ and $$d = 1/abc$$
Also, $$cd = 1/ab, bd = 1/ac, bc = 1/ad$$
The expression can be clubbed together as $$1 + abcd + (a + 1/a) + (b+1/b) + (c+1/c) + (d+1/d) + (ab + 1/ab) + (ac+ 1/ac) + (ad + 1/ad)$$
For any positive real number $$x$$, $$x + 1/x \geq 2$$
Therefore, the least value that $$(a+1/a), (b+1/b)....(ad+1/ad)$$ can take is 2.
$$(a+1)(b+1)(c+1)(d+1) \geq 1 + 1 + 2 + 2 + 2+ 2 + 2 + 2 + 2$$
=> $$(a+1)(b+1)(c+1)(d+1) \geq 16$$
The least value that the given expression can take is 16. Therefore, option C is the right answer.

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