Find the minimum possible value $M$ 
$a,b,c,d,e,f,g$ are non-negative real numbers adding up to $1$. If $$M=\max \{a+b+c, b+c+d, c+d+e, d+e+f,e+f+g\},$$
find the minimum possible value that $M$ can take as $a,b,c,d,e,f,g$ vary.

Please give me some idea to solve this.
 A: Let $M^*$ be the optimal $M$. I will prove the following fact first.

FACT: $M^* \geq \frac{1}{3}$

Proof. There are two cases for the values of $X_1 = a + b + c$ and $X_2 = e + f + g$:


*

*at least one of $X_1$ and $X_2$ is $\geq \frac{1}{3}$

*both $X_1$ and $X_2$ are $< \frac{1}{3}$


For the first case, $M \geq \frac{1}{3}$ because $M \geq \max\{X_1,X_2\}$ by definition. For the second case, we will have $d = 1 - X_1 - X_2 > \frac{1}{3}$, making $M \geq c + d + e > \frac{1}{3}$. In summary, $M^* \geq \frac{1}{3}$ since for $\forall a, b, c, d, e, f, g$, we all have $M \geq \frac{1}{3}$.

Next question: can we reach the lower bound $\frac{1}{3}$? YES! Just let $a = d = g = \frac{1}{3}$ and $b = c = e = f = 0$.
A: Probably a poor answer, but I think the minimum would be when a = d = g = 1/3, the rest being 0, giving M = 1/3.
My trial of thought is that if you were to reduce any of these three values, you'd increase one of the other variables, and the other variables would increase the value of multiple other arguments, resulting in a higher M. Not sure if it's correct, but after some trial and error it's the lowest I could get.
