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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?

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    $\begingroup$ Have you tried any particular values of $x$, $k$? $\endgroup$ – Travis May 11 at 4:47
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    $\begingroup$ The inequality is false for $k=2$. $\endgroup$ – Kavi Rama Murthy May 11 at 4:48
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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 $\ (1-x^k) \ge (1-x)^k.$

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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.

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Without induction you can compare both functions with $y = 1-x$ on $[0,1]$:

$$(1-x)^k =(1-x)(1-x)^{k-1}\stackrel{0\leq 1-x\leq 1}{\leq} 1-x \stackrel{0\leq x\leq 1}{\leq} 1-x^k$$

So, you get the opposite inequality.

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