Reverse Bernoulli's inequality?

Question

For $0<x<1$ and $r>1$, Bernoulli's inequality asserts that $$(1-x)^r\geq 1-rx.$$

Does the reverse inequality hold if we can put a constant in front of $rx$? E.g., $$(1-x)^r\leq 1-\frac{rx}{2}?$$

Answer 1

$$x=\frac12,\,\,r=4\implies (1-x)^r\le 1-\frac{rx}2\iff\left(1-\frac12\right)^4\le1-\frac{4\cdot\frac12}2\iff$$

$$\iff\frac1{16}\le1-\frac22=0$$

so no: it doesn't hold.