Show that $a+b+c=1$ implies $\exists x, y \in \{a-ab, b-bc, c-ca\}$ so that $x \leq \frac{1}{4}$ and $y \geq \frac{2}{9}$ Let $a, b, c$ be three positive real numbers such that $a + b + c = 1$. Prove that among
the three numbers $a − ab, b − bc, c − ca$ there is one which is at most $1/4$ and there
is one which is at least $2/9$
I have proceeded by the AM-GM inequality to prove that \begin{align*}a(1 − a) \leq \frac 14, \qquad b(1 − b) \leq \frac 14, \qquad c(1 − c) \leq \frac 14.\end{align*} Multiplying these we obtain $$abc(1 − a)(1 − b)(1 − c) \leq \frac{1}{4^3}.$$ After that, I could not do much.

Hint 1: For the first inequality, without loss of generality let $a$ be the smallest of the three numbers. Then by the AM-GM, we have
$$a(1-b) \le \left(\frac{a+(1-b)}2\right)^2.$$
Hint 2: For the second inequality, let $a$ be the largest number and $b$ the smallest. Then $a \ge \frac13$ and $1-b \ge \frac23$.
