Prove that $r \in R$ such that $0 \lt r \lt 1$ then $\frac{1}{r (1 - r)} \ge 4$

Question

Prove that $r \in R$ such that $0 \lt r \lt 1$ then $\frac{1}{r (1 - r)} \ge 4$

My method:

Assume that $r \gt 0$, and $1 - r \gt 0$. Hence $r(1 - r) \gt 0$. So, $1 \le 4r (1 - r)$. Hence, $1 \le 4r - 4r^2$. Hence, $4r^2 - 4r + 1 \ge 0$. Thus, $(2r - 1)^2 \ge 0$.

I think I might be missing something but I am not sure.

Answer 1

By GM-HM inequality, $$\sqrt{\frac1r\cdot\frac1{1-r}}\ge\frac{2}{r+(1-r)}$$

Answer 2

Alternately, use the inequality $4ab \leq (a+b)^2$ with $a = r, b = 1-r$ the answer follows immediately.

Answer 3

Alternative: look for the minimum of the function. Taking the derivative with respect to $r$ $$\frac{2x-1}{x^2(1-x)^2}$$ which clearly leads to a minimum in $x=1/2$ (it can be checked with second derivative, for example), and $$\frac{1}{x(1-x)}\ge 4$$