Find the minimum value of expression involving real numbers

Question

Let $n$ positive integer. Find the minimum value of expression:

$$ E=max(\frac {x_1} {1+x_1},\frac {x_2} {1+x_1+x_2}, ... , \frac {x_n} {1+x_1+..+x_n})$$

where $x_1,x_2, .. , x_n$ are real not negative so that $x_1+x_2+ .. +x_n=1$

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My try

For $x_1=1, x_2=0, ...x_n=0$ one gets $E=\frac 1 2$ so $Min \le \frac 1 2$

For $x_1=x_2= ... =x_n=\frac 1 n$ one gets $E=\frac 1 {n+1}$ so $Min \le \frac 1 {n+1}$ (much better)

I suspect the minimum is $\frac 1 {n+1}$ but I cannot prove it (or disprove it).

Update

The minimum cannot be $\frac 1 {n+1}$ because for $n=2, Min=1-\frac{1}{\sqrt2} \lt \frac 1 3$, as @almagest pointed out in a comment.

Answer 1

For $n=2$, the answer is $1-\frac{1}{\sqrt{2}}$, when $x_1=\sqrt{2}-1$ and $x_2=2-\sqrt{2}$.

In general, the answer is $1-\frac{1}{\sqrt[n]{2}}$.

Let's assume $a_0=1$, $a_i=1+x_1+x_2+\dots+x_i$ for $1\leqslant i\leqslant n$. Hence, $a_n=2$ and $a_i\leqslant a_{i+1}$. Then, $\frac{x_i}{1+x_1+x_2+\dots+x_i}=\frac{a_i-a_{i-1}}{a_i}$. Thus,

$$E=\max\{1-\frac{a_0}{a_1},1-\frac{a_1}{a_2},\dots,1-\frac{a_{n-1}}{a_n}\}.$$

Moreover, assume $b_i=\frac{a_{i-1}}{a_i}$ for $i=1,2,\dots, n$. Then, we have $b_1b_2\cdots b_n=\frac{a_0}{a_n}=\frac{1}{2}$, and $\frac{1}{2}\leqslant b_i\leqslant{1}$. Assume, $b_m=\min \{b_1,\dots,b_n\}$, then $b_m\leqslant \frac{1}{\sqrt[n]{2}}$.

$$E=\max\{1-b_1,1-b_2,\dots,1-b_n\}=1-b_m\geqslant 1-\frac{1}{\sqrt[n]{2}}.$$

And the equality holds when $b_1=b_2=\dots=b_n=\frac{1}{\sqrt[n]{2}}$. Note that $b_i=\frac{a_{i-1}}{a_i}=\frac{1+x_1+\dots+x_{i-1}}{1+x_1+\dots+x_{i-1}+x_i}$, we can use induction to prove $x_i=\sqrt[n]{2^i}-\sqrt[n]{2^{i-1}}$ in this case.