Maximum? This shouldn't be tough

Multiplying by $abcd$ , which does not change the expression, we get $(ab+1)(bc+1)(cd+1)(da+1)$ . Now let $x = ab$ and $y = bc$ . Then our expression becomes: $(x+1)(y+1)\left(\frac{1}{x}+1\right)\left(\frac{1}{y}+1\right) = \left(2+x+\frac{1}{x} \right) \left(2+y+\frac{1}{y}\right) = f(x,y)$ Now suppose we have a given value of $x$ and we want to find the value of $y$ which maximizes $f(x,y)$ for that $x$ -value. Let's consider 2 cases:

(Case 1): $x \ge 1$

Then, since the maximum of $a,b$ is 2 and since $ab=x$ , then the minimum of $a,b$ must be $\frac{x}{2}$ . Now $cd = \frac{1}{x}$ . Since the minimum of $c,d$ is $\frac{1}{2}$ , then the maximum of $c,d$ is $\frac{2}{x}$ . Hence, the minimum of $y = bc$ is $\frac{x}{4}$ and the maximum of $y$ is $\frac{4}{x}$ . In other words we have $\frac{x}{4} \le y \le \frac{4}{x}$ .

Now since $\left(2+y+\frac{1}{y}\right)$ is concave upward, the maximum must occur at one of the endpoints (actually both of them, as it turns out). We get $f(x,\frac{4}{x}) = f(x, \frac{x}{4})= \frac{(x+1)^2(x+4)^2}{4x^2}$ . Once again this function is concave upward so the maximum will occur at the endpoints.

So what are the endpoints for $x$ ? Well, since $\frac{1}{2} \le a,b \le 2$ , then $\frac{1}{4} \le x \le 4$ . At both of these endpoints, we get $f(x,\frac{x}{4}) = 25$ .

Therefore the maximum in Case 1 is $25$ . I won't bother writing out Case 2 (where $x < 1$ ) since it is very similar and you get the same maximum.

$a,b,c,d,\frac {1}{a},\frac {1}{b},\frac {1}{c},\frac {1}{d} >0. \\ \text{so, } (a+\frac {1}{a}) + (b+\frac {1}{b}) + (a+\frac {1}{b}) + (a+\frac {1}{b}) \geq \\2 \times \sqrt{(a+\frac{1}{b})(b+\frac{1}{c})(c+\frac{1}{d})(d+\frac{1}{a})}. \text(AM-GM) \\ Otherwise, f(x) = x + \frac {1}{x}. x \in [\frac{1}{2} , 2] \\ f'(x) = 1-\frac {1}{x^2} \\ \text{In this domain,} \\2.5 \geq f(x). \\ 10 \geq (a+\frac {1}{a}) + (b+\frac {1}{b}) + (a+\frac {1}{b}) + (a+\frac {1}{b}) \geq \\ 2 \times \sqrt{(a+\frac{1}{b})(b+\frac{1}{c})(c+\frac{1}{d})(d+\frac{1}{a})}. \\ \text {so } 25 \geq (a+\frac{1}{b})(b+\frac{1}{c})(c+\frac{1}{d})(d+\frac{1}{a}).$

Note several properties of this function:

1. The function is preserved on cyclic permutations of (a, b, c, d)

2. The function can be manipulated to $\left( b + \frac{1}{a} \right) \left( c + \frac{1}{b} \right) \left( d + \frac{1}{c} \right) \left( a + \frac{1}{d} \right)$ by dividing by $(a b c d)^2$ . This means that $f(a, b, c, d) = f(\frac{1}{a}, \frac{1}{b}, \frac{1}{c}, \frac{1}{d})$ .

3. The function is unchanged if $(a, b, c, d)$ is transformed into $(\epsilon a, \frac{b}{\epsilon}, \epsilon c, \frac{d}{\epsilon})$

The constraint $abcd = 1$ along with property 3 means that of the 4 variables, only 2 are free, which will be denoted by $x, y$ . By properties 1 and 2, only two unique orderings exist: $(x, y, \frac{1}{x}, \frac{1}{y})$ and $(x, \frac{1}{x}, y, \frac{1}{y})$ .

$f(x, \frac{1}{x}, y, \frac{1}{y}) = 4 (x + y) (\frac{1}{x} + \frac{1}{y})$

$f(x, y, \frac{1}{x}, \frac{1}{y}) = (x + \frac{1}{y}) (y + \frac{1}{x}) (x + y) (\frac{1}{x} + \frac{1}{y})$

The second value is greater than the first if $(x + \frac{1}{y}) (y + \frac{1}{x}) = 2 + x y + \frac{1}{x y} > 4 \implies x y + \frac{1}{x y} > 2$ . Furthermore, $x y + \frac{1}{x y}$ can be maximized by choosing $x y = 4$ or $\frac{1}{4}$ by virtue of $g(z) = z + \frac{1}{z}$ having a U shape on $(0, \infty)$

Therefore the max value is 25.

You only have 2 possible values for the expression. Either you alternate 2 and 1/2 (2, 1/2, 2, 1/2) or you have them by twos (2, 2, 1/2, 1/2)

Note that (1/2, 2, 1/2, 2) will give you the same values as (2, 1/2, 2, 1/2). Likewise (1/2, 1/2, 2, 2), (2, 1/2, 1/2, 2) and (1/2, 2, 2, 1/2) will give you the same value as (2, 2, 1/2, 1/2)

The possible values would be 16 and 25. Therefore 25 would be the maximum.