Inequalities :::::::::::::::::

Construct a function f(x) = xk - kx + k - 1 for x > 0

If k = 2, then f(x) = x2 - 2x + 1 = (x - 1)2 >= 0

Hence xk - kx + k - 1 >= 0, giving xk - 1 >= k(x - 1)

If k >= 3:

f'(x) = kxk-1 - k

When f'(x) = 0, x = 1

Also f"(x) = k(k - 1)xk-2

Thus f"(1) > 0 since k > 0 and k - 1 > 0

So f(x) attains a minimum at x = 1 with f(1) = 0

Thus xk - kx + k - 1 >= 0, giving xk - 1 >= k(x - 1)

It is not always true.

x is a positive real number. Let x = 0.1

k is a positive integer greater than 1.
When k = 2 :
L.S. = (0.1^2) - 1 = 0.01 - 1 = 0.99
R.S. = 2(0.1 - 1) = 2*(0.9) = 1.8
L.S. < R.S.