Show that a,b,c, and d are positive integers when,
((a^2+1)(b^2+1)(c^2+1)(d^2+1))/abcd≥16
Let
f(a,b,c,d)=(a²+1)(b²+1)(c²+1)(d²+1)/abcd
=(a²+1)/a * (b²+1)/b *(c²+1)/c * (d²+1)/d
What is a positive integer?
It is a member of the set {1,2,3,4.....}
So the smallest positive integer is 1.
Consider the factor
(a²+1)/a
=a+1/a
≥2 when a≥1
therefore
f(a,b,c,d)≥16 when a,b,c,d ∈Z+
To show that a, b, c, and d are positive integers given the expression ((a^2+1)(b^2+1)(c^2+1)(d^2+1))/abcd ≥ 16, we can use proof by contradiction.
Assume that at least one of the variables a, b, c, d is not a positive integer. Without loss of generality, let's assume a is not a positive integer.
Since a is not a positive integer, it must be either a negative integer or a non-integer. Let's consider both cases:
Case 1: a is a negative integer
In this case, a^2 + 1 will always be positive, as the square of a negative integer plus 1 is positive. Therefore, (a^2+1) will be positive, and the expression ((a^2+1)(b^2+1)(c^2+1)(d^2+1))/abcd will also be positive.
Case 2: a is a non-integer
In this case, a^2 + 1 will be greater than 1, as any non-integer squared will be greater than 1. Therefore, (a^2+1) will be greater than 1, and the expression ((a^2+1)(b^2+1)(c^2+1)(d^2+1))/abcd will be greater than the expression (1)(b^2+1)(c^2+1)(d^2+1)/abcd.
Now, let's consider the denominator abcd. Since a is not a positive integer, we know that the product abcd will also not be a positive integer. Therefore, the value of the expression (1)(b^2+1)(c^2+1)(d^2+1)/abcd will be greater than 1, as the denominator abcd is not a positive integer.
Combining the results from Case 2, we have that the expression ((a^2+1)(b^2+1)(c^2+1)(d^2+1))/abcd is greater than 1. This contradicts the given condition that the expression is greater than or equal to 16.
Therefore, our assumption that at least one of the variables a, b, c, d is not a positive integer is incorrect. Hence, a, b, c, and d must all be positive integers.
QED