TITLE: Find min of $\frac{\left( \sum kx_k \right) \left( \sum x^2_k \right)}{\left( \sum x_k \right)^3}$
QUESTION [6 upvotes]: With $n \ge 2$ and $x_1,\ x_2,\ \dots,\ x_n > 0$. Find the minimum of:
$$ M = \frac{(x_1 + 2 x_2 + ...+ nx_n)( x^2_1 + x^2_2 +...+x^2_n)} {\left( x_1 + x_2 +...+ x_n \right)^3}$$

For specific $n$, for example, $n = 3$, I can get: $\min \{M\} = \frac{2}{27}(10 - \sqrt 2)$
But in above generalized problem, how can I solve that ? Can I get the result by using sage or mathematica ?

REPLY [3 votes]: This is not a separate answer, but rather a (too long) comment in response to the answers by
(1) Sergio Parreiras
and (2) user58697 .
To make their answers complete, MAPLE still is not strictly needed but quite useful.
Following steps as pointed out by user58697 results in the following quadratic equation:
$$
n^2(n^2-2n+1)a^2 - 4n^2(n-1)a + (4n^2-2n-2) = 0
$$
Solutions are in a simple closed form:
$$
a = \frac{4n \pm 2\sqrt{2n+2}}{2(n-1)n} \quad \Longrightarrow \quad \min \{M\} = \\
\min{\left(\frac{(2n-\sqrt{2n+2}+2)(2n+\sqrt{2n+2})}{9n(n-1)}\,,\,
           \frac{(2n+\sqrt{2n+2}+2)(2n-\sqrt{2n+2})}{9n(n-1)}\right)}
$$
Giving e.g. for $n=3$:
$$
a = 1\pm\frac{\sqrt{8}}{6}
$$
And for the corresponding function values:
$$
f(a=1-\sqrt{8}/6) = -\frac{2(-4+\sqrt{2})(+3+\sqrt{2})}{27} \approx 0.8454973007 \\
f(a=1+\sqrt{8}/6) = -\frac{2(+4+\sqrt{2})(-3+\sqrt{2})}{27} = \frac{2}{27}(10-\sqrt{2}) \approx 0.6359841806
$$
The latter outcome quite in agreement with the OP's solution.
EDIT. Further simplification leads to this final expression for the minimum:
$$\min\{M\} = \frac{2(n+1)(2n-1)-2\sqrt{2n+2}}{9n(n-1)}$$