A.M and G.M.

ab+bd+ac+cd

=ab+ac+db+dc

=a(b+c)+d(b+c)

=(a+b)(b+c)

also,

a+b+c+d=100

∴ We have to find the highest product of any two nos. who sum to 100 the highest product is 50*50=2500

We assume that both $a+d$ and $b+c$ are non-negative and use AM-GM inequality .

$\begin{aligned} &\dfrac{\left(a+d\right)+\left(b+c\right)}2\geq\sqrt{\left(a+d\right)\left(b+c\right)}\\ &\left(\dfrac{a+b+c+d}2\right)^2 \geq \left(a+d\right)\left(b+c\right) \\ &\left(\dfrac{100}2\right)^2=50^2=2500\geq ab +bd+ac+cd \end{aligned}$

$$

The inequality has equality when $a+d = b+c = 50$ , and both $a+d$ and $b+c$ are non-negative.
Thus, the maximum value of $ab + bd + ac + cd$ is $2500$ , when both $a+d$ and $b+c$ are non-negative

If both $a+d$ and $b+c$ are negative, we have an impossible situation since $a+b+c+d = 100$ .
If either $a+d$ or $b+c$ is negative, we have $ab + bd + ac + cd \leq 0 \leq 2500$ .

$$

Hence, the maximum value of $ab+bd+ac+cd$ is $\boxed{2500}$ for any a,b,c,d which satisfy $a+b+c+d=100$ and $a+d,b+c \in \mathbb{R}$ .

For the max value of ab + bc + cd + da.. take a , b , c , d as 25 each and then we will get the answer as 2500