Let $0<x<1$ and $p<1$ then can I conclude the following : $(1-x)^p<1-px$

Question

Let $0<x<1$ and $p<1$ then can I conclude the following :

$$(1-x)^p<1-px$$

If so how can I show it,

Answer 1

By using Binomial theorem for any index :

$$(1+x)^p=1+px+\frac{p(p-1)}{2!}x^2+\frac{p(p-1)(p-2)}{3!}x^3+\frac{p(p-1)(p-2)(p-3)}{4!}x^4 +\cdots$$

Replace $x\rightarrow -x$

$$(1-x)^p=1-px+\frac{p(p-1)}{2!}x^2-\frac{p(p-1)(p-2)}{3!}x^3+\frac{p(p-1)(p-2)(p-3)}{4!}x^4+\cdots$$ Rearrange :

$$(1-x)^p=1-px-\frac{p(1-p)}{2!}x^2-\frac{p(1-p)(2-p)}{3!}x^3-\frac{p(1-p)(2-p)(3-p)}{4!}x^4 + \cdots $$

Now, since $p\in (0,1)$ and $x>0$,

Clearly ; $$(1-x)^p<1-px$$

Answer 2

If $p \le 0$ this is false. Assume $0 < p < 1$.

A common method for such problems is the first derivative test. Define $$\phi(x) = (1-x)^p - 1 + px,\ 0 \le x \le 1.$$ Then $\phi$ is continuous on $(0,1)$ and $$\phi'(x) = -p(1-x)^{p-1} + p = p \left(1 - (1-x)^{p-1}\right)$$ so that $\phi'(x) < 0$ for $x \in (0,1)$ because $0 < p < 1$. Thus $\phi$ is decreasing on $(0,1)$ so that $$x \in (0,1) \implies \phi(x) < \phi(0) = 0.$$ Thus $(1-x)^p - 1 + px < 0$ for $x \in (0,1)$.

Answer 3

Good answers have already been provided since I posted my comments, but I haven't seen one use my suggestion to evaluate tangent at $0$. We use the derivative of $(1-x)^p$ and the definition of a tangent line along with the concavity of $(1-x)^p$ on $[0,1]$ to derive the desired result

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To accomplish this we first find the slope at $0$ $$\left.\frac{d}{dx} (1-x)^p\right|_{x=0} = \left.-p(1-x)^{p-1}\right|_{x=0} = -p$$ We now note that at $x_0=0$ we have $y_0 = (1-0)^p = 1$
Therefore, our tangent line is the function $$y_0 -p(x-x_0) = 1-px$$ Since $(1-x)^p$ is concave on this interval (check the second derivative) we can say that, $\forall x \in [0,1]$, $$(1-x)^p \le 1-px$$ By checking the endpoints we can likewise say that, $\forall x \in (0,1),$ $$(1-x)^p \lt 1-px \;$$