How to prove relation between $(1-x)^k$ and $(1-x^k)$?

I have been trying to prove or disprove following inequality:

$$(1-x)^k \geq (1-x^k)$$ $$\forall 0 \leq x \leq 1$$ and $$k \in \mathbb{N}$$.

I thought about taking $$\log$$ on both sides and comparing but couldn't reach a conclusion. Does a straight application of binomial series work here? Could you please provide any hint about this?

• Have you tried any particular values of $x$, $k$? – Travis May 11 at 4:47
• The inequality is false for $k=2$. – Kavi Rama Murthy May 11 at 4:48

This is a special case of a more general inequality. Suppose we are given a finite sequence of real numbers $$\ \{x_i\}_{i=1}^k \$$ such that $$\ 0 \le x_i \le 1.\$$ Define the sequence $$\ y_i := 1-x_i\$$ and the finite product of binomials $$P := \prod_{i=1}^k (x_i + y_i) = \prod_{i=1}^k 1 = 1.$$ Using distributivity of multiplication over addition, we can expand the product $$\ 1 = P = \prod_{i=1}^k x_i +\ ...\ + \prod_{i=1}^k y_i \ge \prod_{i=1}^k x_i + \prod_{i=1}^k y_i\$$ because the missing $$\ 2^k-2\$$ products are all non-negative. Now we can specialize to $$\ x = x_i \$$ for all $$\ i\$$ and the resulting inequality is $$\ 1 \ge x^k + (1-x)^k,\$$ or equivalently, it becomes $$\ (1-x^k) \ge (1-x)^k.$$
The reverse inequality is true and you can prove it by induction: if the reverse inequality holds for $$k$$ then $$(1-x)^{k+1}\leq (1-x^{k})(1-x) \leq 1-x^{k+1}$$ as you can easily check.
Without induction you can compare both functions with $$y = 1-x$$ on $$[0,1]$$:
$$(1-x)^k =(1-x)(1-x)^{k-1}\stackrel{0\leq 1-x\leq 1}{\leq} 1-x \stackrel{0\leq x\leq 1}{\leq} 1-x^k$$