Find the Upper Bound

Titu's lemma (Cauchy):

$\sum \frac{n^2}{d} \ge \frac{(\sum n)^2}{\sum d}$

Hence,

$\frac{(x_1+2x_2+\dots+2014x_{2014})^2}{x_1^2+x_2^2+\dots+x_{2014}^2}\le1^2+2^2+\dots+2014^2$ whose digit sum is $\boxed{38}$ .

By Titu's lemma (see here ) or Cauchy-Schwarz (multiply by the RHS on the right side), $\sum_{i=1}^{2014}\dfrac{i^2x_i^2}{x_i^2}=\dfrac{1^2x_1^2}{x_1^2}+\dfrac{2^2x_2^2}{x_2^2}+\cdots+\dfrac{2014^2x_{2014}^2}{x_{2014}^2}\ge \dfrac{(x_1+2x_2+\cdots+2014x_{2014})^2}{x_1^2+x_2^2+\cdots+x_{2014}^2}$ Then, $N=\sum_{i=1}^{2014}\dfrac{i^2x_i^2}{x_i^2}=\sum_{i=1}^{2014} i^2=\dfrac{(2014)(2015)(4029)}{6}=2725088015$ which has digit sum of $\boxed{38}$ .

Motivation: The expression looks suspiciously like Titu's lemma, and also if I see a square that is less than something, I think Cauchy-Schwarz.

It's not 2014 yet!...

Since we want a maximum value of the LHS, assume that $x_1 \leq x_2 \leq x_3 \leq \cdots \leq x_{2014}$ . We can assume that because if $x_i > x_j$ for $i < j$ , then we can optimize the expression further by switching values for $x_i$ and $x_j$ .

The LHS looks a lot like a rearrangement of Cauchy-Schwarz / Titu's Lemma. On the RHS = N, we will have an expression made up of $2014$ fractions, with numerators chosen from the set $\{x_1^2, (2x_2)^2, (3x_3)^2, \cdots, (2014x_{2014})^2\}$ and denominators chosen from the set $\{x_1^2, x_2^2, x_3^2, \cdots, x_{2014}^2\}$ . We wish to minimize $N$ , and so by the rearrangement inequality we know that it is minimized when the greatest numerator is paired up with the greatest denominator. Thus, we come to

$\frac{(x_1 + 2x_2 + \cdots + 2014x_{2014})^2}{x_1^2 + x_2^2 + \cdots + x_{2014}^2} \leq \displaystyle \sum_{n = 1}^{2014} \frac{n^2 x_n^2}{x_n^2} = \displaystyle \sum_{n = 1}^{2014} n^2 = 2 725 088 015$

It is a simple application of the Cauchy-Shwarz Inequality .

$(\sum_{i=1}^{i=2014} ix_i)^2 \le \sum_{i=1}^{i=2014} i^2 \sum_{i=1}^{i=2014} x_i^2$ This can be rearranged to the inequality given in the question and value of $N$ is $N=\sum_{i=1}^{i=2014} i^2=\frac{2014\times 2015\times 4029}{6}=2725088015$ Thus sum of digits is 38

Notice by Cauchy-Schwarz, $(x_1^2 + x_2^2 + \cdots + x_{2014}^2)(1^2 + 2^2 + \cdots + 2014^2) \ge (x_1 + 2x_2 + \cdots + 2014x_{2014})^2$ Therefore, $\begin{aligned}\dfrac{(x_1 + 2x_2 + \cdots + 2014x_{2014})^2}{(x_1^2 + x_2^2 + \cdots + x_{2014}^2)} &\le& (1^2 + 2^2 + \cdots + 2014^2) \\ &=& \dfrac{2014 \cdot 2015 \cdot 4029}{6}\\ &=& 2725088015\end{aligned}$ Equality occurs when $x_1 = 1, \ x_2 = 2, \ \cdots, x_{2014} = 2014$ . Therefore, $N = 2725088015$ so our answer is $\boxed{38}$ . $\blacksquare$

By the look of this problem, it's pretty clear that it wants to be killed by the Cauchy-Schwarz Inequality. Letting $\{a_i\}_{i=1}^{2014}=x_i$ and $\{b_i\}_{i=1}^{2014}=i$ , by Cauchy we get:

$(x_1+2x_2+...+2014x_{2014})^2 \le (1^2+2^2+3^2+...+2014^2)(x_1^2+x_2^2+...+x_{2014}^2)$

Rearranging yields:

$\frac{(x_1+2x_2+...+2014x_{2014})^2}{x_1^2+x_2^2+...+x_{2014}^2} \le 1^2+2^2+3^2+...+2014^2=N$

At this point (actually at the very beginning of the problem lol =P), the solution becomes trivial...

$N=1^2+2^2+3^2+...+2014^2=\frac{(2(2014)+1)(2014+1)(2014)}{6}=272508815$ .

Then the sum of the digits is: $2+7+2+5+8+8+1+5=\boxed{38}$ .

as it is applicable to all real series let x1=1 therefore,it becomes {1^2+2^2+3^2...................+2014^2014}^2 divided by 1^2+2^2+3^2+4^2.......................+2014^2014 therefore the expression becomes1^2+2^2+3^2+4^2........2014^2014====which can be calculated by the formulae for the sum of squares of consecutive integers======n(n+1)(2n+1)/6.the sum of digits will come 38

By Cauchy-Schwarz, we have $\left[\sum_{i=1}^{2014} i \cdot x_i\right]^2 \leq \sum_{i=1}^{2014} i^2 \cdot \sum_{i=1}^{2014} x_i^2$ , which rearranges to yield $\frac{\left[\sum_{i=1}^{2014} i \cdot x_i\right]^2}{\sum_{i=1}^{2014} x_i^2} \leq \sum_{i=1}^{2014} i^2 = 2725088015$ .Hence, the sum of the digits of $N$ is $\boxed{38}$

By Cauchy Swartz inequality,

[ (x1)^2 + (x2)^2 + ......... + (x2014)^2 ] [ 1 +4 +9 +16 + ...........] ≥ [x1 + 2(x2) +.........+ 2014(x2014)]^2

So the given expression in the question in less than or equal to 1+4 +9 +........+2014^2 and we can get N by using sum of squares of first n naturals and then we can find the sum of digits as 38