Minimum of the maximum Let $a, b,c,d,e,f,g$  seven  nonnegative real  numbers  with  the  sum  equal  to  1. Let  considered  the sums $ a+b+c$, $b+c+d$, $c+d+e$, $d+e+f$, $e+f+g$. What  is  the  minimum  of  the  maximum   of  those  sums  when  we  consider   all  the  possibilities?I  have  no  idea.  One  hint, please?

The language is obfuscating but:
Suppose $a=b=c=0$ then $d+e+g = 1$.  $d+e+g$ and that is the maximum of the sums.  But that's pretty high maximum.  What if we don't have three elements equal to $0$.  Then any sum of three numbers will not include a fourth possible none zero term.  So the sum of any three can not be as high as $1$.  So in that case the maximum will be less than $1$. 
So what's the minimum the maximum can be.  That's what they are asking.  
Let me explain WOLOG (without loss of generality).  The letters we use don't matter.  One of those numbers must be the least number of the seven.  Let's call that one $a$.  Without loss of generality we might as well assume that $a$ is the least element because ... some number must be and we can just relabel.   Likewise we can assume $b$ is the second least and so on.
So WOLOG $a \le b \le c \le d\le e\le g$. And so the maximum of the sums is $d+e+g$.  What is the least that $d+e+g$ can be if $a \le b \le c \le d\le e\le g$?
That is what is really being asked.
Now $1 = a+b+c+d+e+f+g \ge (a+b+c) + (a+b+c) = 2(a+b+c)$ so $a+b+c \le \frac 12$.
So $d+e+g = 1 - (a+b+c) \ge \frac 12$.  So $d+e+g$ will be at least $\frac 12$ so $\frac 12$ is the absolute minimum $d+e+g$ can be.
But can $d + e+g = \frac 12$?  Well, it can if $a+b+c= \frac 12$ but can $a+b+c = \frac 12$? Well it can if $a = b = c = \frac 16$.  And then $d+e+g$ can equal $\frac 12$ if $d=e=g=\frac 16$.
So the minimum of the maximum is $\frac 12$.
It'll take a bit more work to show that the only way $d+e +g = \frac 12$ is if all terms are equal to $\frac 16$ but we weren't asked to show that.
But here's how:
If $g > \frac 16$ and $d+e+g = \frac 12$ then $d+e < \frac 12 - \frac 16 = \frac 13$.  So $d+e < \frac 13$ and $2d = d + d \le d+e <\frac 13$ so $d < \frac 16$.
So $a\le b \le c \le d < \frac 16$ so $a + b + c < \frac 16 + \frac 16 + \frac 16 = \frac 16$.  But then $\frac 12 = d+e+g = 1-(a+b+c) > \frac 12$ which is a contradiction.
So $d+e+g =\frac 12 \iff g = \frac 16$.  $1 = a+b+c+d+e+g \ge a+b+c+d+g+g \ge a+b+c+g+g+g \ge a+b+g+g+g+g \ge a+g+g+g+g+g \ge g+g+g+g+g+g = 1$.  Thus $ a+b+c+d+e+g = a+b+c+d+g+g = a+b+c+g+g+g = a+b+g+g+g+g = a+g+g+g+g+g = g+g+g+g+g+g$ so $a=b=c=d=e=g=\frac 16\iff d+e+g =\frac 12$.
