Let $a,b,c$ are positive reals satisfying $\frac{2}{5} \leq c \leq \min\{a,b\}$ ; $ac \geq \frac{4}{15}$ and $bc \geq \frac{1}{5} .$ Prove that $$ \frac{1}{a}+\frac{2}{b}+\frac{3}{c} \leq 13$$Equality holds when $ a=\frac{2}{3},b=\frac{1}{2},c=\frac{2}{5} .$

interesting problem！

I think I over-expected from the problem... Outline: Basically, make two cases on whether $\min\{a,b\}$ is $a$ or $b$. If $a \ge b$, then picking $c = \frac{2}{5}$ works directly with no problems[look at post #1]. For the other case, just write $c = \frac{2}{5} + \epsilon$ for some $\epsilon > 0$. Then solve the other inequality cases, to get upper bounds on $a,b$. Reciprocating and adding, would give you that $\epsilon = 0$ is the best choice for attaining a maximum of $13$ which is impossible to attain since $b \ge a$. That's it!