amparvardi wrote: Find the minimum value of \[\max(a + b + c, b + c + d, c + d + e, d + e + f, e + f + g)\] subject to the constraints (i) $a, b, c, d, e, f, g \geq 0,$ (ii)$ a + b + c + d + e + f + g = 1.$ Let $m$ be the maximum for given $a,...,g$. Then \[3m\ge (a + b + c)+(c + d + e)+( e + f + g)=1+c+e\ge 1,\] thus $m\ge\frac 13$. This minimum can be achieved by $a=d=g=\frac 13,b=c=e=f=0.$

amparvardi wrote: Find the minimum value of \[\max(a + b + c, b + c + d, c + d + e, d + e + f, e + f + g)\] subject to the constraints (i) $a, b, c, d, e, f, g \geq 0,$ (ii)$ a + b + c + d + e + f + g = 1.$ Let $K=\max\{a+b+c, b+c+d, c+d+e, d+e+f, e+f+g\}$ then we have $\begin{array}{l} K \ge a, \\ K \ge a + b, \\ K \ge a + b + c, \\ K \ge b + c + d, \\ K \ge c + d + e, \\ K \ge d + e + f, \\ K \ge e + f + g, \\ K \ge f + g, \\ K \ge g. \\ \end{array}$ adding them together, we get $9K\ge3(a+b+c+d+e+f+g)$ i.e. $K\ge\frac{1}{3}$, when $a=d=g=\frac{1}{3},b=c=e=f=0,K=\frac{1}{3}$, so $K_{\min}=\frac{1}{3}$.

See also here: http://www.artofproblemsolving.com/Forum/viewtopic.php?f=38&t=59737 http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&t=60144