3. (CAN) 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) $a+b+c+d+e+f+g=1$.
3. Denote $\max (a+b+c, b+c+d, c+d+e, d+e+f, e+f+g)$ by $p$. We have  $$ (a+b+c)+(c+d+e)+(e+f+g)=1+c+e \leq 3 p $$  which implies that $p \geq 1 / 3$. However, $p=1 / 3$ is achieved by taking $(a, b, c, d, e, f, g)=(1 / 3,0,0,1 / 3,0,0,1 / 3)$. Therefore the answer is $1 / 3$. Remark. In fact, one can prove a more general statement in the same way. Given positive integers $n, k, n \geq k$, if $a_{1}, a_{2}, \ldots, a_{n}$ are nonnegative real numbers with sum 1 , then the minimum value of $\max _{i=1, \ldots, n-k+1}\left\{a_{i}+\right.$ $\left.a_{i+1}+\cdots+a_{i+k-1}\right\}$ is $1 / r$, where $r$ is the integer with $k(r-1)<n \leq k r$.