Prove or disprove $\sum\limits_{1\le i < j \le n} \frac{x_ix_j}{1-x_i-x_j} \le \frac18$ for $\sum\limits_{i=1}^n x_i = \frac12$($x_i\ge 0, \forall i$)

Question

Problem 1: Let $x_i \ge 0, \, i=1, 2, \cdots, n$ with $\sum_{i=1}^n x_i = \frac12$. Prove or disprove that $$\sum_{1\le i < j \le n} \frac{x_ix_j}{1-x_i-x_j} \le \frac18.$$

This is related to the following problem:

Problem 2: Let $x_i \ge 0, \, i=1, 2, \cdots, n$ with $\sum_{i=1}^n x_i = \frac12$. Prove that $$\sum_{1\le i<j\le n}\frac{x_ix_j}{(1-x_i)(1-x_j)}\le \frac{n(n-1)}{2(2n-1)^2}.$$

Problem 2 is in "Problems From the Book", 2008, Ch. 2, which was proposed by Vasile Cartoaje.
See: Prove that $\sum_{1\le i<j\le n}\frac{x_ix_j}{(1-x_i)(1-x_j)} \le \frac{n(n-1)}{2(2n-1)^2}$

Background:

I proposed Problem 1 when I tried to find my 2nd proof for Problem 2.

It is not difficult to prove that $$\frac{1}{(2n-1)^4} + \frac{16n^2(n-1)^2}{(2n-1)^4}\cdot \frac{x_ix_j}{1-x_i-x_j} \ge \frac{x_ix_j}{(1-x_i)(1-x_j)}.$$ (Hint: Use $\frac{x_ix_j}{(1-x_i)(1-x_j)}= 1 - \frac{1}{1 + x_ix_j/(1-x_i-x_j)}$ and $\frac{1}{1+u} \ge \frac{1}{1+v} - \frac{1}{(1+v)^2}(u-v)$ for $u = x_ix_j/(1-x_i-x_j)$ and $v=\frac{1}{4n(n-1)}$. Or simply $\mathrm{LHS} - \mathrm{RHS} = \frac{(4x_ix_jn^2 - 4x_ix_j n + x_i + x_j - 1)^2}{(2n-1)^4(1-x_i-x_j)(1-x_i)(1-x_j)}\ge 0$.)

To prove Problem 2, it suffices to prove that $$\frac{1}{(2n-1)^4}\cdot \frac{n(n-1)}{2} + \frac{16n^2(n-1)^2}{(2n-1)^4}\sum_{1\le i < j \le n} \frac{x_ix_j}{1-x_i-x_j} \le \frac{n(n-1)}{2(2n-1)^2} $$ or $$\sum_{1\le i < j \le n} \frac{x_ix_j}{1-x_i-x_j} \le \frac18.$$

For $n=2, 3, 4$, the inequality is true.

For $n=5, 6$, numerical evidence supports the statement.

Any comments and solutions are welcome and appreciated.

Answer 1

Write $p_i = 2x_i$ and note that $\sum_i p_i = 1$. Then

\begin{align*} 1 + \sum_i \frac{p_i^2}{1 - p_i} &= \sum_i \frac{p_i}{1 - p_i} \\ &= \sum_{i,j} \frac{1}{2} \left( \frac{1}{1 - p_i} + \frac{1}{1 - p_j} \right) p_i p_j \\ &\geq \sum_{i,j} \left( \frac{2}{2-p_i-p_j} \right) p_i p_j. \tag{by AM–HM} \end{align*}

Rearranging this inequality, we get

$$ 1 \geq \sum_{i \neq j} \frac{2p_i p_j}{2 - p_i - p_j} = 8 \sum_{i < j} \frac{x_i x_j}{1 - x_i - x_j},$$

completing the proof.

Answer 2

Partial proof with hint :

Let us consider the inequality for $x,y>0$ and $x,y\leq 0.5$ :

$$l\left(x,y\right)=g\left(2xy+\frac{4}{3}xy\left(x-y\right)^{2}\right)-\frac{x^2y^2}{1-x^2-y^2}\ge 0$$

Where $g\left(x\right)=\frac{\frac{x^{2}}{4}}{1-x}$

As $g$ is convex on $[0,0.5]$ we use Jensen-Mercer inequality

We need to show :

$$\frac{n\left(n-1\right)}{2}\left(h\left(x,y\right)+h\left(a,b\right)-g\left(f\left(x,y\right)+f\left(a,b\right)-d\right)\right)\le\frac{1}{8}\tag{I}$$

Where $f\left(x,y\right)=2xy+\frac{4}{3}xy\left(x-y\right)^{2},h\left(x,y\right)=g\left(f\left(x,y\right)\right)$ and $f(a,b)\leq d \leq f\left(x,y\right)$

Now we use the inverse function of $f(x)$ with a positive value or :

$$r\left(x\right)=2\sqrt{x^{2}+x}-2x$$

So $(I)$ is true with the constraint (where we have used Karamata's inequality)

$$f\left(x,y\right)+f\left(a,b\right)\le r\left(\frac{\frac{1}{8}2}{n\left(n-1\right)}+g\left(f\left(x,y\right)+f\left(a,b\right)-d\right)\right)+r(0)$$

Answer 3

An incomplete approach:

$\sum\limits_{1\le i < j \le n} A_i A_j=\frac{1}{2}[(\sum_{k=1}^{n} A_k )^2=\sum_{k=1}^{n} A_k^2]=$

Using integral representation, we can write $\sum\limits_{1\le i < j \le n} \frac{x_ix_j}{1-x_i-x_j} =\sum\limits_{1\le i < j \le n} \int_{0}^{1} x_i x_j t^{-x_i-x_j} dt=\int_{0}^{1} dt\sum\limits_{1\le i < j \le n} x_i t^{-x_i} ~x_j t^{-x_j}= \frac{1}{2} [(\sum_{k=1}^{n} \int_{0}^{1} x_k t^{-x_k})^2+\sum_{k=1}^{n} \int_{0}^{1} x_k^2 t^{-2x_k}]dt= \frac{1}{2} \left[\frac{x_k^2}{(1-x_k)^2}-\frac{x_k^2}{1-2x_k}\right]$

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