logic with mathematics

Suppose K = (1+a)(1+b)(1+c)(1+d) ---(1)

Taking a,b,c,d common from each of the terms

K = a(1+1/a)×b(1+1/b)×c(1+1/c)×d(1+1/d)

=> K = abcd(1+1/a)(1+1/b)(1+1/c)(1+1/d)

=> K = (1+1/a)(1+1/b)(1+1/c)(1+1/d) -----(2) since abcd = 1

Multiplying (1) and (2) we get,

K^2 = (1+a)(1+b)(1+c)(1+d)(1+1/a)(1+1/b)(1+1/c)(1+1/d)

=> K^2 = (1+a)(1+1/a)(1+b)(1+1/b)(1+c)(1+1/c)(1+d)(1+1/d) -----(3)

Now (1+a)(1+1/a) = (1+a+1/a+1)=(2+a+1/a)

a+1/a ≥ 2 (AM ≥ GM)

=> 2+a+1/a ≥ 4

Hence, (1+a)(1+1/a) ≥ 4

Similarly

(1+b)(1+1/b) ≥ 4

(1+c)(1+1/c) ≥ 4

(1+d)(1+1/d) ≥ 4

Multiplying the 4 terms we get

(1+a)(1+1/a)(1+b)(1+1/b)(1+c)(1+1/c)(1+d)(1+1/d) ≥ 4×4×4×4

=> K^2 ≥ 256 (from (3))

=> K ≥ 16

Thus the minimum value of K is 16.

Since $a$ is positive real, we can apply AM-GM inequality as $1+a \ge 2\sqrt a$ and equality occurs when $a=1$ . Similarly,

$(1+a)(1+b)(1+c)(1+d) \ge 2\sqrt a \cdot 2\sqrt b \cdot 2\sqrt c \cdot 2\sqrt d = 16 \sqrt{abcd} = \boxed{16}$

Equality occurs when $a=b=c=d=1$ .