Don't just guess

We' re going to apply AM>=GM inequality: a+1>=2sqrt(a); b+1>=2sqrt(b); c+1>=2sqrt(c); d+1>=2sqrt(d); (a+1)(b+1)(c+1)(d+1)>=16sqrt(abcd); (a+1)(b+1)(c+1)(d+1)>=16sqrt(1); (a+1)(b+1)(c+1)(d+1)>=16(1), we get the minimum value when
(a+1)(b+1)(c+1)(d+1)=16

Note that we can use AM - GM Inequality since it is explicitly given that $a$ , $b$ , $c$ , and $d$ are positive reals. Expanding $(1 + a)(1 + b)(1 + c)(1 + d)$ gives

$1^4 - [- (a + b + c + d)] × 1^3 + (ab + ac + ad + bc + bd + cd) × 1^2 - [- (abc + abd + acd + bcd)] × 1 + abcd$ ( From Vieta's Identities )

or

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

We can now use the AM - GM Inequality to find the minimum value of the above expression. Thus,

$\frac{1 + a + b + c + d + ab + ac + ad + bc + bd + cd + abc + abd + acd + bcd + abcd}{16}$ $\ge$ $\sqrt[16]{(1 × a × b × c × d × ab × ac × ad × bc × bd × cd × abc × abd × acd × bcd × abcd)}$ $=$ $\sqrt[16]{(abcd)^8} =$ $\sqrt[16]{1^8} =$ $1$

Moreover, we get

$1 + a + b + c + d + ab + ac + ad + bc + bd + cd + abc + abd + bcd + abcd$ $\ge$ $16 × 1 = 16$

Hence, the minimum value is $16$ which is then obtained when $1 = a = b = c = d = ab = ac = ad = bc = bd = cd = abc = abd = acd = bcd = abcd$ . From here, it follows that $a = b = c = d = 1$ .

Note: I'm sorry for the tedious solution. I just wanted to be clear that's why I wrote it that way.

suppose a=b=c=d=1 and u can get the answer :)

An intuitive exploration:

First, let's limit our exploration to just a and b.

In order for ab = 1, a and b must be reciprocals.

if a = 1, then b = 1

So (1+1)(1+1) = 2 * 2 = 4

if a = 2 then b = 1/2

So (1+2)(1+1/2) = 9/2 = 4.5

If a = 3 then b = 1/3

So (1+3)(1+1/3) = 16/3 = 5.33333...

As we use smaller fractions, the reciprocals are inversely larger, and the products of the sum of each with 1 are also larger, so using 1 for all of the variables gives the minimum product.

(1+1)(1+1)(1+1)(1+1) = 2 * 2 * 2 * 2 = 16

The minimum point (a,b,c,d) that minimaze f(a,b,c,d)=(1+a)(1+b)(1+c)(1+d) under the constraint abcd=1 also minimaze ln[f(a,b,c,d)] under the same constraint. let H(a,b,c,d)=ln[f(a,b,c,d)] and also we have g(a,b,c,d)=abcd=1 now using lagrange multiplier, that is "Gradient(H(a,b,c,d))=λ Gradient(g(a,b,c,d)) and we get that λ =a/(1+a) , λ =b/(1+b) , λ =c/(1+c) and λ =d/(1+d) Which implies a=b=c=d but because abcd=1 hence a=b=c=d=1 therefore min[f]=f(1,1,1,1)=16