Using titu's lemma , we get that $ \frac{1}{2014 -1} .K \ge 1 \Longrightarrow K \ge 2013 $

it is too trivial assuming , $\sum x_i=1$ the sequences $x_i$'s and $(1-x_i)^{-1}$'s are sorted in a same way so applying Chebyshev's Ineq, $P=\sum \frac {x_i^2}{1-x_i} \geq \frac 1n (\sum x_i^2).(\sum \frac {1}{1-x_i})$ $ \implies P\geq \frac {1}{n^2} (\sum x_i)^2. \frac {n^2}{n-\sum x_i}$ by repeated application of CS implies , $P\geq \frac {1}{n-1}$ plugg $n=2014 \cdots $ Also jenesen works here too !

Trivial application of Cauchy-Schwarz. Basically we want to find minimum value of $S=\sum\dfrac{x_j^2}{1-x_j}$ subject to $\sum x_j=1$. So $(\sum\dfrac{x_j^2}{1-x_j})(\sum {1-x_j})\geq (\sum{x_j})^2=1\iff S\geq\dfrac{1}{2013}\iff K=2013$.

During the test,I did this in two ways.One using A.M-G.M and the other uses Cauchy.I shall present the A.M-G.M one. Note that $\frac{1}{1-x_j}-x_j=1+\frac{{x_j}^2}{1-x_j}$ Summing over all $j$ this yeilds $\sum_{j=1}^{2014}{\frac{1}{1-x_j}}-2015=\sum_{j=1}^{2014}{\frac{{x_j}^2}{1-x_j}}$. By A.M-G.M we get $(\prod_{i=1}^{2014}{(1-x_j)})^{\frac{1}{2014}} \le \frac{2013}{2014}$ Again using A.M-G.M and the last line we get $\sum_{j=1}^{2014}{\frac{1}{1-x_j}} \ge 2014{\frac{1}{\prod_{i=1}^{2014}{(1-x_j)}}}^{\frac{1}{2014}} \ge \frac{{2014}^2}{2013}$ Hence from the first equality we get $\sum_{j=1}^{2014}{\frac{{x_j}^2}{1-x_j} \ge \frac{{2014}^2}{2013}-2015=\frac{1}{2013}}$ Hence $K=2013$.Equality holds when every $x_i$ is equal.Hence $K$ cannot be made lesser.

I think there is something wrong with the question. For example, consider any of the x's to be equal to 0.9. Then, x^2/1-x = .81/.1 = 8.1 the sum for all of the other x's should be positive, so when we take the final sum, it will be K(8.1 + something). In this case, K doesn't have to be greater than 1. Repeat this, and use 0.99, 0.999, etc. I believe we'll find that the value of K continuously decreases and tends to zero, and there is no minimum for k.

That value for $K$ should work for all allowed values of those variables; you just exhibit some particular set of values, for which even a lesser $K$ will work ...

I wrote the answer 2013, but they didn't specify that it has to work for every possible arrangement. It just said the smallest k such that the inequality CAN be satisfied. there is thus no bound on K.

Just (typical) carelessness from the problem setters in wording the statement.

By C-S , $\sum_{j=1}^{2014}\frac{x_j^2}{1-x_j} \ge\frac{(\sum_{j=1}^{2014}x_j)^2}{2014-\sum_{j=1}^{2014}x_j}=\frac{1}{2013}$, Equality holds when $x_1=x_2=\cdots =x_{2014}=\frac{1}{2014}$, hence $K_{min}=2013$ .

$\Sigma_{j=1}^{2014}(1+x_j)$=2015 $\implies \Sigma_{j=1}^{2014}(\frac {1-x_j^2} {1-x_j})$=2015 $\implies \Sigma_{j=1}^{2014}(\frac {1} {1-x_j})$=2015 + $\Sigma_{j=1}^{2014}(\frac {x_j^2} {1-x_j})$ Now by AM-HM $\frac {\Sigma_{j=1}^{2014}(\frac {1} {1-x_j})} {2014} \ge {2014} {\Sigma_{j=1}^{2014}(1-x_j)}$ $\implies \Sigma_{j=1}^{2014} (\frac {1} {1-x_j}) \ge \frac {2014^2} {2013}$ $\implies$ 2015 + $\Sigma_{j=1}^{2014}(\frac {x_j^2} {1-x_j}) \ge \frac {2014^2} {2013}$ $\implies \Sigma_{j=1}^{2014}(\frac {x_j^2} {1-x_j}) \ge \frac {1} {2013}$ So K=2013

From http://www.artofproblemsolving.com/Forum/viewtopic.php?f=51&t=585619 we can see that the answer is $\boxed{K=2013}$. Perhaps India copied my question!

sorry for bumping the thread again but everybody is giving their different respnses but actually there is no need,simply apply titu and that should do it.

Commander_Anta78 wrote: sorry for bumping the thread again but everybody is giving their different respnses but actually there is no need,simply apply titu and that should do it. Bro... Giving more than one solution which are not similar is not a sin, I guess... LOL

Smathematician wrote: Commander_Anta78 wrote: sorry for bumping the thread again but everybody is giving their different respnses but actually there is no need,simply apply titu and that should do it. Bro... Giving more than one solution which are not similar is not a sin, I guess... LOL Oh I am sorry