Here is my question: Suppose that a and b are real numbers with 0 < b < a. Prove

Question

Here is my question: Suppose that a and b are real numbers with 0 < b < a. Prove that if n is a positive integer, then an - bn <= nan-1(a - b).
I'm fine with every step up until the inductive step and I don't know how to compare the inequality in a way that makes sense. Hope you can help!

Here is my question: Suppose that a and b are real numbers with 0 < b < a. Prove that if n is a positive integer, then an - bn <= nan-1(a - b).
I'm fine with every step up until the inductive step and I don't know how to compare the inequality in a way that makes sense. Hope you can help!

Explanation / Answer

S(n) = a^n - b^n < [n*a^(n-1)](a-b).


For induction, we need three steps to check for:

S(n = 0, 1, ni) smallest number inside the interval
Assume S(n = n) is true S(n = n+1) This is because we want to prove that for the first number in the interval it is true. From this we can assume safely that for some n value we will get a true answer (since n is the number just chosen technically.) The kicker is to prove that for the next number after n (n+1) is true.


The First Test:
n = 1
S(n=1) = a^(1) - b^(1) < [(1)a^(1-1)](a-b)
           = a-b < (a-b) (CHECK)
since 0 < b < a, (a-b) > 0 We Assume S(n = n) is True (as given in the inequality) Test For S(n = n+1): S(n+1) = a^(n+1) - b^(n+1) < [(n+1)a^(n)]*[(a-b)]
(a)*a^(n) - (b)*b^(n) < [n*a^(n) + a^(n)]*[(a-b)]
(a)*a^(n) - (b)*b^(n) < [(n*a)a^(n) - (n*b)a^(n) + (a)a^(n) - (b)a^(n)] (n*a)a^(n) - (n*b)a^(n) > (a)*a^(n) - (b)*b^(n)    since a > b, and n is a positive integer.
(a)a^(n) - (b)a^(n) > (a)*a^(n) - (b)*b^(n)    since a > b, and n is a positive integer.

Therefore,

(n*a)a^(n) - (n*b)a^(n) > (a)*a^(n) - (b)*b^(n)    since a > b, and n is a positive integer.
(a)a^(n) - (b)a^(n) > (a)*a^(n) - (b)*b^(n)    since a > b, and n is a positive integer.

Therefore,
(a)*a^(n) - (b)*b^(n) < [(n*a)a^(n) - (n*b)a^(n) + (a)a^(n) - (b)a^(n)] (CHECK)
In other words, each terms on the right is larger than the left hand side of the inequality, therefore their sum is also greater than the left hand side of the inequality and therefore we have proven what we wanted to prove:
a^n - b^n < [n*a^(n-1)](a-b). (QED) I Hope this helps!
(a)*a^(n) - (b)*b^(n) < [(n*a)a^(n) - (n*b)a^(n) + (a)a^(n) - (b)a^(n)] (CHECK)
In other words, each terms on the right is larger than the left hand side of the inequality, therefore their sum is also greater than the left hand side of the inequality and therefore we have proven what we wanted to prove:
a^n - b^n < [n*a^(n-1)](a-b). (QED) I Hope this helps!
a^n - b^n < [n*a^(n-1)](a-b). (QED) I Hope this helps!

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