Find the least of the $\lambda$ such this inequality hold? 
Give postive integer  $n$,  find the least real numbers $\lambda$, and $0<a_{i}<2^i,\forall i=1,2,\cdots,n$. have
  $$\sum_{i=1}^{n}\sum_{j=1}^{n}\{a_{i}a_{j}\}\le \lambda  \sum_{i=1}^{n}\{a_{i}\}$$ where $\{x\}=x-[x]$.

This inequality I have try sometimes,and I can't solve it. and I  have heard my country's There are  no people who solved this problem
Now it is said the answer is
$$\lambda_{\min}=2^{n+1}-n-2+\dfrac{n+(2^{n+1}-n-2)(2^n-1)}{\sqrt{(2^n-1)^2+1}}$$
 A: I can prove that the least required $\lambda$ belongs to the interval  $$\left[2^{n+2}-2n-4+n(\sqrt{(2^n-1)^2+1}-(2^n-1)),2^{n+2}-n-4\right].$$
Given numbers $0<a_{i}<2^i$, for each $i$ and $j$ we have 
$$\{a_{i}a_{j}\}=\{([a_i]+\{a_i\})([a_j]+\{a_j\}) \}=$$
$$\{[a_i][a_j]+[a_i]\{a_j\}+[a_j]\{a_i\}+\{a_i\}\{a_j\}\}=$$
$$\{[a_i]\{a_j\}+[a_j]\{a_i\}+\{a_i\}\{a_j\}\}\le $$
$$\{[a_i]\{a_j\}\}+\{[a_j]\{a_i\}\}+\{\{a_i\}\{a_j\}\}\le $$
$$ (2^i-1)\{a_j\}+(2^j-1)\{a_i\}+\{a_i\}\{a_j\}.$$
Put $S=\sum_{i=1}^{n}\{a_i\}$. We have 
$$\sum_{i=1}^{n}\sum_{j=1}^{n}\{a_{i}a_{j}\}\le 
\sum_{i=1}^{n}\sum_{j=1}^{n} (2^i-1)\{a_j\}+(2^j-1)\{a_i\}+\{a_i\}\{a_j\}=$$ 
$$2S\sum_{j=1}^{n} (2^j-1)+S^2=(2^{n+2}-2n-4)S+S^2\le(2^{n+2}-n-4)S.$$
Thus $\lambda\le 2^{n+2}-n-4$.
On the other hand, given $n$, pick a positive $\varepsilon$ such that $(2^{n+1}-2)\varepsilon+\varepsilon^2<1$ and put $a_i=2^i-1+\varepsilon$ for each $i$. Then $\{a_i\}=\varepsilon$ and $\{a_ia_j\}=(2^i+2^j-2)\varepsilon+\varepsilon^2$. We can easily caclulate that the left hand side of the inequality equals $(n2^{n+2}-2n^2-4n)\varepsilon+n^2\varepsilon^2$ and the left hand side equals $n\lambda\varepsilon$, so  $\lambda\ge 2^{n+2}-2n-4+n\varepsilon$. Since $\varepsilon$ can be chosen arbitrarily close to $\sqrt{(2^n-1)^2+1}-(2^n-1)$, we have $\lambda\ge 2^{n+2}-2n-4+n(\sqrt{(2^n-1)^2+1}-(2^n-1)).$
