This is the general formula for total resistance of resistors in parallel:
$${1 \over R_T} = {1 \over R_1} + {1 \over R_2} + ... + {1 \over R_n}$$
Now let's assume we have a resistor in this circuit that has the least resistance and give that resistance a value of \(a\). That means that all the other resistors have a coefficient, \(k_i\) where \(k_i \ge 1\). Now our total resistance in the circuit is:
$${1 \over R_T} = {1 \over a} + {1 \over k_1a} + ... + {1 \over k_na}$$
Now we make a common denominator to get:
$${1 \over R_T} = {(k_1k_2...k_{n-1}k_n + k_2k_3...k_{n-1}k_n + ... + k_1k_2...k_{n-1}k_n) \over k_1k_2...k_{n-1}k_na}$$
But we are not looking for \(1 \over R_T\). We are looking for \(R_T\). So we invert the equation to get:
$${R_T} = \frac{k_1k_2...k_{n-1}k_na}{k_1k_2...k_{n-1}k_n + k_2k_3...k_{n-1}k_n + ... + k_1k_2...k_{n-1}k_n}$$
Notice that the first term of the denominator is the same as the coefficient of the numerator. Since the rest of the terms of the denominator is at least one because all values of k_i are at least one we can replace those terms with the variable \(x\) to get:
$${R_T} = \frac{k_1k_2...k_{n-1}k_n}{k_1k_2...k_{n-1}k_n + x}{a}$$
Since \(x \ge 1\) the coefficient of \(a\ \lt 1\) and thus \(R_T\ \lt a\), which is the resistor of least resistance in the parallel circuit.
$${R_T \lt a}$$