If a + b + c + d = 2, then the maximum value of (1 + a)(1 + b)(1 + c)(1 + d) is

Given, a + b + c + d = 2

We know that, AM $$\ge\ $$ GM

$$\Rightarrow$$ Arithmetic mean of (1 + a),(1 + b),(1 + c),(1 + d) $$\ge\ $$ Geometric mean of (1 + a),(1 + b),(1 + c),(1 + d)

$$\Rightarrow$$ $$\frac{\left(1+a\right)+\left(1+b\right)+\left(1+c\right)+\left(1+d\right)}{4}\ge\left[\ \left(1+a\right)\left(1+b\right)\left(1+c\right)\left(1+d\right)\right]^{\frac{1}{4}}$$

$$\Rightarrow$$ $$\frac{4+a+b+c+d}{4}\ge\left[\left(1+a\right)\left(1+b\right)\left(1+c\right)\left(1+d\right)\right]^{\frac{1}{4}}$$

$$\Rightarrow$$ $$\frac{4+2}{4}\ge\left[\left(1+a\right)\left(1+b\right)\left(1+c\right)\left(1+d\right)\right]^{\frac{1}{4}}$$

$$\Rightarrow$$ $$\frac{6}{4}\ge\left[\ \left(1+a\right)\left(1+b\right)\left(1+c\right)\left(1+d\right)\right]^{\frac{1}{4}}$$

$$\Rightarrow$$ $$\left[\left(1+a\right)\left(1+b\right)\left(1+c\right)\left(1+d\right)\right]^{\frac{1}{4}}\le\ \frac{3}{2}$$

$$\Rightarrow$$ $$\left(1+a\right)\left(1+b\right)\left(1+c\right)\left(1+d\right)\le\ \left(\frac{3}{2}\right)^4$$

$$\Rightarrow$$ $$\left(1+a\right)\left(1+b\right)\left(1+c\right)\left(1+d\right)\le\ \frac{81}{16}$$

$$\therefore\ $$Maximum value of (1 + a)(1 + b)(1 + c)(1 + d) = $$\frac{81}{16}$$

Hence, the correct answer is Option D