For $n=3$ it is equivalent to Vasc's Inequality (posted in Crux) and for $n=4$ hungktn mentioned it here (post #32): http://www.artofproblemsolving.com/Forum/viewtopic.php?f=51&t=36873 also, the solution given by Zhaobin (post #23) on that thread also work here, with some spices

Inequality is equivalent to: $\sum_{1\le i < j\le n}\frac{n}{n-a_{i}a_{j}} \le \frac{n^2}{2}$ $\Leftrightarrow \sum_{1\le i < j\le n}\frac{a_ia_j}{n-a_{i}a_{j}} \le \frac{n}{2}$ Using Am-Gm and Cauchy-Schwarz, we have: $ \sum_{1\le i < j\le n}\frac{a_ia_j}{n-a_{i}a_{j}} \leq \sum_{1\le i < j\le n}\frac{2a_ia_j}{2n-a_{i}^2 -a_{j}^2 } \\ \leq \frac{1}{2} \sum_{1\le i < j\le n}\frac{(a_i+a_j)^2}{(n-a_{i}^2)+(n-a_{j}^2)} \leq \frac{1}{2} \sum_{1\le i < j\le n} \left( \frac{a_i^2}{n-a_{j}^2}+\frac{a_j^2}{n-a_{i}^2} \right)=\frac{n}{2}$ Equality holds when $a_1=...=a_n=1$. $\blacksquare$

Generalization http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&t=473078

MathUniverse wrote: Equality holds when $a_1=...=a_n=1$. $\blacksquare$ Equality also holds when $a_1=...=a_n=-1$ ..

Yes.. I haven't seen that $a_1,...,a_n$ are real numbers.

Hang on, this is NOT a cyclic inequality, take a look at the summation \[ \sum_{1 \le j <j\le n}\frac{1}{n-a_ia_j} =\frac{1}{n-a_1a_2}+\frac{1}{n-a_2a_3}+\ldots+\frac{1}{n-a_{n-1}a_n} \]

http://www.mmjp.or.jp/competitions/APMO/files/apmo2012_sol.pdf

Firstly notice that proving inequality to the positives will force to be true for reals too. It is equivalent to the inequality \[ \sum_{1\le i < j\le n}\frac{1-a_i a_j}{n-a_i a_j}\ge\ 0 \] Let $a_i a_j=x_{ij}$ Which is equivalent to \[ \sum_{1\le i < j\le n}\frac{1-(x_ij)^2}{(n-x_ij)(1+x_ij)}\ge\ 0 \] Now we use Chebyshev's Inequality Suppose $x_{ij}\ge x_{kl}$ Then we get $1-(x_{ij})^2\le1-(x_{kl})^2$ and $\frac{1}{(n-x_{ij})(1+x_{ij})}\le \frac{1}{(n-x_{kl})(1+x_{kl})}$ since the last inequality is equivalent to $(x_{ij}-x_{kl})(n-1-x_{ij}-x_{kl})\ge 0$ Case 1(here we assume $i,j,k,l$ are all different Last inequlaity is true since $n-1\ge \frac{n}{2}\ge \frac{(a_i)^2+(a_j)^2+(a_k)^2+(a_l)^2}{2} \ge x_{ij}+x_{kl}$ Case 2(we assume $j=l$ Then we need $n-1\ge a_l (a_i+a_k)$ suppose $n\ge 4$ adn converse hold $ n-1 \le a_l (a_i+a_k) \le \frac {(a_l)^2 +(a_i+a_k)^2}{2} \le \frac {(a_l)^2 }{2}+(a_i)^2+(a_k)^2$ if and $ n-1 \le a_l (a_i+a_k) \le \frac {(a_l)^2 +(a_i)^2+(a_l)^2+(a_k)^2}{2} \le {(a_l)^2 + \frac{(a_i)^2+(a_k)^2}{2}}$ Summing last to inqualities we get $2(n-1)\le \frac{3}{2}((a_i)^2+(a_l)^2+(a_k)^2)$ but it force $ $2(n-1)\le \frac{3}{2}n$ from which $n<4$. So by Chebyshevs \sum_{1\le i < j\le n}\frac{1-(x_{ij})^2}{(n-x_{ij})(1+x_{ij})}\ge\ \frac {(n(n-1)-\sum_{1\le i<j\le n}(a_i a_j)^2)(\sum_{1\le i<j\le n}\frac{1}{(n-x_{ij})(1+x_{ij})})}{n(n-1)}

This works, right?

MathUniverse wrote: Inequality is equivalent to: $\sum_{1\le i < j\le n}\frac{n}{n-a_{i}a_{j}} \le \frac{n^2}{2}$ $\Leftrightarrow \sum_{1\le i < j\le n}\frac{a_ia_j}{n-a_{i}a_{j}} \le \frac{n}{2}$ Using Am-Gm and Cauchy-Schwarz, we have: $ \sum_{1\le i < j\le n}\frac{a_ia_j}{n-a_{i}a_{j}} \leq \sum_{1\le i < j\le n}\frac{2a_ia_j}{2n-a_{i}^2 -a_{j}^2 } \\ \leq \frac{1}{2} \sum_{1\le i < j\le n}\frac{(a_i+a_j)^2}{(n-a_{i}^2)+(n-a_{j}^2)} \leq \frac{1}{2} \sum_{1\le i < j\le n} \left( \frac{a_i^2}{n-a_{j}^2}+\frac{a_j^2}{n-a_{i}^2} \right)=\frac{n}{2}$ Equality holds when $a_1=...=a_n=1$. $\blacksquare$ Nice proof! Was the Cauchy-Schwarz application there the Cauchy-Schwarz Inequality in Engel form? antimonyarsenide wrote: This works, right?

Fixed the $LaTeX$. Nice solution, and how did you think of using jensen on the function $f(x)=\frac{1}{n-x}$? Thanks!

I thinks it's convex since $f(x)=(n-x)^{-1} \rightarrow f'(x)=(n-x)^{-2} \rightarrow f''(x)=2(n-x)^{-3} \ge 0$

Wow this inequality is really sharp. Here is a completely ridiculous solution for $n\ge4$ using the tangent parabola method : We will instead prove that $\sum_{1\le i<j\le n}\frac{a_ia_j}{n-a_ia_j}\le\frac n2$, which is equivalent as $\sum\frac{a_ia_j}{n-a_ia_j}=\sum\left(\frac n{n-a_ia_j}-1\right)=-\binom n2+n\sum\frac1{n-a_ia_j}$. Note that $a_ia_j\le\frac{a_i^2+a_j^2}2\le\frac{a_1^2+a_2^2+\cdots+a_n^2}2=\frac n2$ for all $i,j$. The key idea is that for $t\le\frac n2$ we have that $\frac t{n-t}\le\frac{2t^2+(n-4)t+1}{(n-1)^2}$. Indeed, cross multiplying and collecting terms gives $-2t^3+(n+4)t^2-(2n+2)t+n\ge0$, which is equivalent to $(t-1)^2(n-2t)\ge0$. Hence, we have that $\sum\frac{a_ia_j}{n-a_ia_j}\le\frac2{(n-1)^2}\sum a_i^2a_j^2+\frac{n-4}{(n-1)^2}\sum a_ia_j+\frac1{(n-1)^2}\sum 1$. Applying the bounds $\sum a_i^2a_j^2\le\frac{n-1}{2n}(a_1^2+a_2^2+\cdots+a_n^2)^2=\frac{n(n-1)}2$ and $\sum a_ia_j\le\frac{n-1}2(a_1^2+a_2^2+\cdots+a_n^2)=\frac{n(n-1)}2$, which respectively follow from $\sum(a_i^2-a_j^2)^2\ge0$ and $\sum(a_i-a_j)^2\ge0$, we have that $\sum\frac{a_ia_j}{n-a_ia_j}\le\frac2{(n-1)^2}\cdot\frac{n(n-1)}2+\frac{n-4}{(n-1)^2}\cdot\frac{n(n-1)}2+\frac1{(n-1)^2}\cdot\frac{n(n-1)}2=\frac n2$, as desired.

MathUniverse wrote: Yes.. I haven't seen that $a_1,...,a_n$ are real numbers. However，it’s not difficult to see that x_i have the same sign

It's easy to see that we can assume WLOG that the $a_i$'s are all positive. Now, by dividing each $a_i$ by $\sqrt{n}$, we will solve the equivalent problem where $\sum a_i^2 = 1$ and we wish to show that $\sum_{i \neq j} \frac{1}{1-a_ia_j} \leq \frac{n^2}{2},$ i.e., $\sum_{i \neq j} \frac{a_ia_j}{1-a_ia_j} \leq \frac{n}{2}.$ From the Tangent Line Trick, we can obtain that: $$\frac{x}{1-x} \leq \frac{2n^2}{(n-1)^2}x^2 + \frac{n^2-4n}{(n-1)^2}x + \frac{1}{(n-1)^2},$$ whenever $x \in [0, \frac{1}{2}],$ where we got this by setting a quadratic to be tangent to the curve at the point $x = \frac{1}{n}$ (when equality happens at $a_i \equiv \frac{1}{\sqrt{n}}$) and to attain equality at $x = \frac{1}{2}$ ($a_ia_j \leq \frac12$ by AM-GM). When $n \geq 4$, we know that all of the coefficients of the above quadratic are nonnegative, and so using the facts that $\sum_{i \neq j} a_i^2a_j^2 \leq \frac{n-1}{2n}$ and $\sum_{i \neq j} a_ia_j \leq \frac{n-1}{2}$ (both obvious using $\sum a_i^2 = 1$), we get that the maximal value of $\sum_{i \neq j} \frac{a_ia_j}{1-a_ia_j}$ occurs exactly when equality holds in these two inequalities, i.e. when $a_i a_j = \frac{1}{n}$ for all $i \neq j$. Therefore, this maximal value is just $\frac{n(n-1)}{2} * \frac{\frac{1}{n}}{1 - \frac{1}{n}} = \frac{n}{2},$ as desired. When $n =2,$ the problem is trivial. So therefore, let's turn our attention to the $n= 3$ case. In this case, let our variables be $a, b, c$. Then, using the variables above, we know that: $$\sum_{cyc} \frac{ab}{1-ab} \leq \frac92 (a^2b^2 + b^2c^2 + c^2a^2) - \frac34 (ab + bc + ca) + \frac34,$$ and so in this case we wish to show that $\frac92(a^2b^2 + b^2c^2 + c^2a^2) - \frac34(ab + bc + ca) \leq \frac34,$ which upon using the condition $a^2 + b^2 + c^2 = 1$, scaling, then homogenizing is equivalent to: $$6(a^4+b^4+c^4) + (a+b+c)^2(a^2+b^2+c^2) \geq 5(a^2+b^2+c^2)^2.$$Expanding this and dividing by $2$, we want that: $$a^4+b^4+c^4 + a^3b+ab^3 + b^3c+bc^3 + c^3a+ca^3 +abc(a+b+c) \geq 4(a^2b^2 + b^2c^2 + c^2a^2),$$ which miraculously is equivalent to: $$ab(a-b)^2 + bc(b-c)^2 + ca(c-a)^2 + (a+b+c)(a^3+b^3+c^3 + 3abc - a^2b-ab^2-b^2c-bc^2-c^2a-ca^2),$$which we know is $\geq 0$ by the Trivial Inequality and Schur's Inequality!

First , we notice that $\sum_{1\leq i < j \leq n } 1 = \sum_{1\leq i < j \leq n } \frac{a_i^2+a_j^2}{2} =\binom{n}{2}$. By AM-GM $ \sum_{1\leq i < j \leq n } \frac{1}{n-a_ia_j} \leq \sum_{1\leq i < j \leq n } \frac{1}{n-\frac{a_i^2+a_j^2}{2}}$. Then divide $\frac{n}{2}$ into $(n-1)\cdot \binom{n}{2}$. We just need to prove $ \sum_{1\leq i < j \leq n }(\frac{1}{n-1}- \frac{1}{n-\frac{a_i^2+a_j^2}{2}})= \sum_{1\leq i < j \leq n } \frac{1-\frac{a_i^2+a_j^2}{2}}{(n-1)(n-\frac{a_i^2+a_j^2}{2})} \geq 0$. $\Leftrightarrow \sum_{1\leq i < j \leq n } \frac{1-\frac{a_i^2+a_j^2}{2}}{(n-\frac{a_i^2+a_j^2}{2})} \geq 0$. Claim:$ \sum_{1\leq i < j \leq n } (n-\frac{a_i^2+a_j^2}{2})(1-\frac{a_i^2+a_j^2}{2}) \geq 0$. Proof:$LHS = \sum_{1\leq i < j \leq n } [n -(n+1) \frac{a_i^2+a_j^2}{2} +(\frac{a_i^2+a_j^2}{2})^2] \geq 0 $ $\Leftrightarrow \sum_{1\leq i < j \leq n } (\frac{a_i^2+a_j^2}{2})^2 \geq \frac{n(n-1)}{2} $. $\sum_{1\leq i < j \leq n } (\frac{a_i^2+a_j^2}{2})^2 = \frac{(n-2)\sum_{i=1}^n a_i^4 +(\sum_{i=1}^n a_i^2)^2}{4} \geq \frac{(\frac{n-2}{n} +1)n^2}{4}= \frac{n(n-1)}{2} $. By Chauchy $ \sum_{1\leq i < j \leq n } (n-\frac{a_i^2+a_j^2}{2})(1-\frac{a_i^2+a_j^2}{2}) \cdot \sum_{1\leq i < j \leq n } \frac{1-\frac{a_i^2+a_j^2}{2}}{(n-\frac{a_i^2+a_j^2}{2})} \geq (\sum_{1\leq i < j \leq n } (1-\frac{a_i^2+a_j^2} {2} ) )^2 =0$ So $ \sum_{1\leq i < j \leq n } \frac{1-\frac{a_i^2+a_j^2}{2}}{(n-\frac{a_i^2+a_j^2}{2})} \geq 0$ as desired.