Find min of $\frac{\left( \sum kx_k \right) \left( \sum x^2_k \right)}{\left( \sum x_k \right)^3}$

Question

With $n \ge 2$ and $x_1,\ x_2,\ \dots,\ x_n > 0$. Find the minimum of: $$ M = \frac{(x_1 + 2 x_2 + ...+ nx_n)( x^2_1 + x^2_2 +...+x^2_n)} {\left( x_1 + x_2 +...+ x_n \right)^3}$$

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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 ?

Answer 1

Let $$f(\mathbf{x})=\frac{(x_1 + 2 x_2 + ...+ nx_n)( x^2_1 + x^2_2 +...+x^2_n)} {\left( x_1 + x_2 +...+ x_n \right)^3}$$.

1. $f$ is homegeous of degree zero: $f(\alpha\mathbf{x})=f(\mathbf{x})$ for all $\alpha>0$.
2. Without any loss of generality we can assume $\sum_{i=1}^n x_i=1$.
3. $\min\limits_{\mathbf{x}} (x_1 + 2 x_2 + ...+ nx_n)( x^2_1 + x^2_2 +...+x^2_n)$ subject to $\sum_{i=1}^n x_i=1$.
4. First order condition wrt $x_i$: $i\,(x^2_1 + x^2_2 +...+x^2_n)+2x_i(x_1 + 2 x_2 + ...+ nx_n)=\lambda$. FOC
5. You can use Mathematica (see Reduce) to solve the system of FOC.
6. But probability better to try to simplify FOCs before using Mathematica: $$\dfrac{i+1}{i}(\lambda-2x_{i+1}((x_1 + 2 x_2 + ...+ nx_n)=\lambda-2x_{i}((x_1 + 2 x_2 + ...+ nx_n)$$
7. For $i=N$ you can not simplify FOC. Don't forget to include $\sum x_i=1$ as well. You need $N+1$ equations since you are also solving for $\lambda$.
8. For $N=4$, using Maple I got :

x[1] = (1/2)RootOf(10_Z^2-20*_Z+9),

x[2] = 1/6+(1/6)RootOf(10_Z^2-20*_Z+9),

x[3] = -(1/6)RootOf(10_Z^2-20*_Z+9)+1/3

Or [x[1] = .3418861170, x[2] = .2806287057, x[3] = .2193712943]

of course that x[4] = 1-x[1]-x[2]-x[3]

Answer 2

Maple is not needed. Subtracting FOCs for $i$ and $i+1$ gives $\sum x_k^2 + 2(x_i - x_{i+1}) \sum kx_k = 0$. Since neither sum depends on $i$, you may conclude that $x_i - x_{i+1}$ is a constant, that is $x$s form an arithmetic progression, $x_k = a + bk$.

Let $S_m = \sum k^m$, i.e. ($S_0 = n, S_1 = \frac{n(n-1)}{2}, S_2 = \frac{n(n+1)(2n+1)}{6}$)

Now, $$\sum kx_k = \sum k(a + bx_k) = S_1a + S_2b$$

$$\sum x_k^2 = \sum a^2 + 2abk + b^2k^2 = S_0a^2 + 2S_1ab + S_2b^2$$

That is, $$f(x) = (S_1a + S_2b)(na^2+2S_1ab + S_2b^2)$$ And since $\sum x_k = 1$, $$S_0a + S_1b = 1$$

Substituting $b = \frac{1 - S_0a}{S_1}$ shows that $f$ is a cubic polynomial of $a$, so $\frac{df}{da} = 0$ is a quadratic equation.

Answer 3

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: $$\min\{M\} = \frac{2(n+1)(2n-1)-2\sqrt{2n+2}}{9n(n-1)}$$