How prove this inequality with $\{x_{i}\}$

Question

if $n>3$ be give positive integers,let $x_{i}>0$ and $x_{i}\notin Z,i=1,2,\cdots,,n$,such $x_{1}+x_{2}+\cdots+x_{n}=n$,find in closed form the best constant $C_{1}(n),C_{2}(n)$ such $$C_{1}(n)\le \sum_{i=1}^{n}\dfrac{x_{i}\{x_{i}\}}{1+x_{i}}\le C_{2}(n)$$

where $\{x\}=x-[x]$

Now I found the best $C_{1}(n)=\dfrac{1}{n+1},$

take $x_{1}=x_{2}=\cdots=x_{n-2}=1-\varphi,x_{n-1}=2-\varphi,x_{n}=(n-1)\varphi,$ where $\varphi\to 0^{+}$,

and $C_{2}(n)=\dfrac{n-1}{2}+\dfrac{2}{3}$

take$x_{1}=x_{2}=\cdots=x_{n-1}=\varphi,x_{n}=n-(n-1)\varphi$,where $\varphi\to 0^{+}$,

But I can't prove following inequality $$\dfrac{1}{n+1}\le \sum_{i=1}^{n}\dfrac{x_{i}\{x_{i}\}}{1+x_{i}}\le \dfrac{n-1}{2}+\dfrac{2}{3}$$

Answer 1

Given a set $\{x_1,x_2,\cdots ,x_n\}$ such that $\sum x_i=n$, it is clear that at least two of $\{x_i\}$s , say $x_j$ and $x_k$, are not equal. Hence the summation $$S=\sum {x_i\{x_i\}\over 1+x_i}$$ contains a term $${x_j\{x_j\}\over 1+x_j}+{x_k\{x_k\}\over 1+x_k}$$. WLOG, assume $x_j>x_k$. Since both $x_j$ and $x_k$ are non-integers, there exists a deviation $\epsilon$ with $|\epsilon|$ sufficiently small such that $\lfloor x_j+\epsilon\rfloor=\lfloor x_j\rfloor$ and $\lfloor x_k+\epsilon\rfloor=\lfloor x_k\rfloor$. Also we know that $${x_j\{x_j\}\over 1+x_j}+{x_k\{x_k\}\over 1+x_k}{={x_j^2-x_j\lfloor x_j\rfloor\over 1+x_j}+{x_k^2-x_k\lfloor x_k\rfloor\over 1+x_k}\\=x_j+x_k-2-\lfloor x_j\rfloor-\lfloor x_k\rfloor+{1+\lfloor x_j\rfloor\over 1+x_j}+{1+\lfloor x_k\rfloor\over 1+x_k}}$$By transforming $(x_j,x_k)$ to $(x_j+\epsilon,x_k-\epsilon)$ for sufficiently small $|\epsilon|$, the term $x_j+x_k-2-\lfloor x_j\rfloor-\lfloor x_k\rfloor$ remains unchanged and we obtain $${(x_j+\epsilon)\{x_j+\epsilon\}\over 1+x_j+\epsilon}+{(x_k-\epsilon)\{x_k-\epsilon\}\over 1+x_k-\epsilon}-\left({x_j\{x_j\}\over 1+x_j}+{x_k\{x_k\}\over 1+x_k}\right)\\={1+\lfloor x_j+\epsilon\rfloor\over 1+x_j+\epsilon}+{1+\lfloor x_k-\epsilon\rfloor\over 1+x_k-\epsilon}-{1+\lfloor x_j\rfloor\over 1+x_j}-{1+\lfloor x_k\rfloor\over 1+x_k}\\=\epsilon \left( {1+\lfloor x_k\rfloor\over (1+x_k)(1+x_k-\epsilon)}-{1+\lfloor x_j\rfloor\over (1+x_j)(1+x_j+\epsilon)} \right)$$It is impossible that the term inside the parenthesis holds zero for every $\epsilon$, hence the difference can be positive or negative for proper values of $\epsilon$ which concludes that the set $\{x_i\}$ cannot obtain supremum nor infimum.

The key results is that each of $x_i$ must be infinitesimally close to an integer. Based on this, let $$x_i=k_i+\theta_i$$ where $|\theta_i|$ is infinitesimally small. Consider all of the non-negative integer solutions of $$k_1+\cdots+k_n=n$$. First consider the solution where $k_1=k_2=\cdots=k_n=1$. Hence all $x_i$s are close to $1$ from above or below. If exactly $p$ numbers of $x_i$s are close to $1$ from below and the rest from above we obtain $$S=\sum {x_i\{x_i\}\over 1+x_i}=\sum_{x_i=1^-} {1\over 2}$$therefore $${1\over 2}<S<{n-1\over 2}$$

For all other solutions, we have at least one $k_i=0$ which must be $k_i=0^+$. Let $$x_j=k_j^\pm\quad,\quad j\ne i$$ where some of $x_j$s are $k_j^+$ and the rest are $k_j^-$. Note that we must have at least one $x_j$ equal to $k_j^-$ for $j\ne i$. The summation is maximum when all $x_j$s are $k_j^-$, which leads to $$S=\sum {k_j\over k_j+1}$$and is maximum when the solution $(2,1,1,\cdots ,1,0)$ is considered which leads to $$S={n-2\over 2}+{2\over 3}$$ The minimum is obtained when all but one $x_u$ are $k_j^+$. Call the remaining $x_u$ as $m\ne 0$. Hence $$S={m\over m+1}$$ which is minimum if $m=1$. Therefore the following inequality is obtained:

$${1\over 2}<\sum {x_i\{x_i\}\over 1+x_i}<{n-2\over 2}+{2\over 3}$$