Question: Suppose $0<a<b<c<1$, find the minimum of $\max(b-a,c-b,1-c)$ if $b \ge 2a$ or $ a+b \le 1 $.
Here is my try:
Since $ \begin{cases} \max(b-a,c-b,1-c) \ge b-a \\ \max(b-a,c-b,1-c) \ge c-b \\ \max(b-a,c-b,1-c) \ge 1-c \\ \end{cases} $,we have $ 3 \times \max(b-a,c-b,1-c)\ge 1-a$. Then, $$\max(b-a,c-b,1-c) \ge \cfrac{1-a}{3} \qquad (*) $$ with equality if and only if $b-a=c-b=1-c$, or, equivalently,$\begin{cases} b=\cfrac{2a+1}{3} \\c=\cfrac{a+2}{3} \end{cases}$.
Since $b \ge 2a$ or $ a+b \le 1 $, it follows that $a \le \cfrac{2}{5}$ when the equality in $(*)$ holds.
Hence $\max(b-a,c-b,1-c) \ge \cfrac{1-a}{3} \ge \cfrac{1}{5}$,where the last equality holds iff $a=\cfrac{2}{5},b=\cfrac{3}{5},c=\cfrac{4}{5}$.
Am I right?
Is this solution correct and clear enough? Or, are there points therein where it lacks in accuracy or clarity of logic?