# Find the minimnum of the $\max{(2x_{1}+x_{2},2x_{2}+x_{3},\cdots,2x_{n-1}+x_{n})}$

Let $n$ is give postive integers,and for $x_{i}\ge 0$,such $$x_{1}+x_{2}+\cdots+x_{n}=1$$ find $I$ minimum of the value $$I=\max{(2x_{1}+x_{2},2x_{2}+x_{3},\cdots,2x_{n-1}+x_{n})}$$

I try to find the $$2x_{1}+x_{2}\le I$$ $$2x_{2}+x_{3}\le I$$ $$\cdots$$ $$2x_{n-1}+x_{n}\le I$$ add all $$(n-1)I\ge 3(x_{1}+\cdots+x_{n})-x_{1}-2x_{n}=3-x_{1}-2x_{n}$$ but $x_{1}+x_{n}$ maximum is?

• If you add the inequalities, shouldn't you get $(n-1)I\geq 3-x_1-2x_n$? Commented Aug 24, 2016 at 14:41
• @Batominovski,Thanks ,I have edit Commented Aug 24, 2016 at 14:45

Assume that $n\geq 2$. Observe that, as $x_1,x_2,\ldots,x_n\geq0$ (in particular, $x_{n-1}\geq 0$), we have \begin{align} S&:=\sum_{i=1}^{n-2}\,\left(\frac{2^i-(-1)^i}{3}\right)\,2^{n-2-i}\,\left(2x_{i}+x_{i+1}\right)+2^{n-2}\left(2x_{n-1}+x_n\right) \\ &=\left(\frac{2^n-(-1)^n}{3}\right)\,x_{n-1}+2^{n-2}\,\sum_{i=1}^{n}\,x_i\geq 2^{n-2}\,\sum_{i=1}^n\,x_i=2^{n-2}\,. \end{align} On the other hand, \begin{align} S&\leq I\left(\sum_{i=1}^{n-2}\,\left(\frac{2^i-(-1)^i}{3}\right)\,2^{n-2-i}+2^{n-2}\right) \\&= I\,\left(2^{n-2}+(n-2)\frac{2^{n-2}}{3}-\frac{(-1)^n}{3}\,\sum_{i=1}^{n-2}(-2)^{n-2-i}\right) \\ &=\frac{I}{3}\left((n+1)2^{n-2}-(-1)^n\,\sum_{i=0}^{n-3}\,(-2)^i\right) \\ &=\frac{I}{3}\Biggl((n+1)2^{n-2}-(-1)^n\,\left(\frac{1-(-2)^{n-2}}{3}\right)\Biggr) \\ &=\frac{I}{3^2\cdot 2^2}\big((3n+3)\,2^n-4\cdot(-1)^n+2^n\big) \\ &=\frac{I}{3^2\cdot 2^2}\big((3n+4)\, 2^n-4(-1)^n\big)\,. \end{align}\,. Ergo, $$I\geq \frac{3^2\cdot 2^2\cdot S}{(3n+4)\, 2^n-4(-1)^n}\geq \frac{9\cdot 2^n}{(3n+4)\, 2^n-4(-1)^n}\,.$$ Therefore, $$I\geq \frac{9\cdot 2^n}{(3n+4)\, 2^n-4(-1)^n}=:m_n\,.$$ The equality holds if and only if $2x_i+x_{i+1}=m_n$ for all $i=1,2,\ldots,n-1$ and $x_{n-1}=0$, which then yields the unique minimizing point: $$\left(x_1,x_2,\ldots,x_n\right)=\frac{m_n}{3}\,\Biggl(1-\frac{1}{(-2)^{n-2}},1-\frac{1}{(-2)^{n-3}},1-\frac{1}{(-2)^{n-4}},\ldots,0,3\Biggr)\,.$$ For instances, $m_2=1$, $m_3=\frac{2}{3}$, $m_4=\frac{4}{7}$, and $m_5=\frac{8}{17}$ with the following respective minimizing points: $(0,1)$, $\left(\frac{1}{3},0,\frac{2}{3}\right)$, $\left(\frac{1}{7},\frac{2}{7},0,\frac{4}{7}\right)$, and $\left(\frac{3}{17},\frac{2}{17},\frac{4}{17},0,\frac{8}{17}\right)$.
• For $n=2$, your minimum value gives us $\frac32$, but I think $(x_1,x_2)=(0,1)$ is a feasible solution that give a smaller value of $1$, isn't it? Commented Aug 24, 2016 at 16:37