let $a_{1},a_{2},\cdots,a_{n}\ge 0$,and such $a_{1}+a_{2}+\cdots+a_{n}=1$.

Find this follow minimum $$I=a^2_{1}+a^2_{2}+\cdots+a^2_{n}-2a_{1}a_{2}-2a_{2}a_{3}-\cdots-2a_{n-1}a_{n}-2a_{n}a_{1}$$

My try:since $$a^2_{1}+a^2_{2}+\cdots+a^2_{n}\ge a_{1}a_{2}+a_{2}a_{3}+\cdots+a_{n-1}a_{n}+a_{n}a_{1}$$ this is true because $$\Longleftrightarrow \dfrac{1}{2}[(a_{1}-a_{2})^2+(a_{2}-a_{3})^2+\cdots+(a_{n-1}-a_{n})^2+(a_{n}-a_{1})^2]\ge0$$ so \begin{align*} &a^2_{1}+a^2_{2}+\cdots+a^2_{n}-2a_{1}a_{2}-2a_{2}a_{3}-\cdots-2a_{n-1}a_{n}-2a_{n}a_{1}\\ &\ge a^2_{1}+a^2_{2}+\cdots+a^2_{n}-2(a^2_{1}+a^2_{2}+\cdots+a^2_{n})\\ &=-(a^2_{1}+a^2_{2}+\cdots+a^2_{n}) \end{align*} if we use Cauchy-Schwarz inequality,we have $$(a^2_{1}+a^2_{2}+\cdots+a^2_{n})(1+1+\cdots+1)\ge (a_{1}+a_{2}+\cdots+a_{n})^2=1$$

But this is not usefull.

and Then I can't, yesterday I have ask this problem:How find this inequality minimum $\sum_{i=1}^{n}a^2_{i}-2\sum_{i=1}^{n-1}a_{i}a_{i+1}$,

and the Ewan Delanoy use nice method to solve it.I don't know this problem have this nice methods to solve it too(maybe can use previous question methods?but I can't use it.) and this problem have someone have research? if no,I think this is nice problem.

and Now I guess This problem when $a_{i}=\dfrac{1}{n}$ then $I$ is minimum?and the minimum is $$I_{min}=-\dfrac{1}{n}?$$

Thank you

  • $\begingroup$ For $n \geq 7,$ you can use the answer to the previous question while demanding that $a_1$ be one of the entries set to $0,$ so it is guaranteed that the value $-1/6$ can be achieved, improving on $-1/n.$ I imagine that is best for $n \geq 6. $ $\endgroup$ – Will Jagy Nov 13 '13 at 5:33
  • $\begingroup$ Hello,How can you use the previous question? can you post you full solution?Thank you $\endgroup$ – math110 Nov 13 '13 at 6:27

UPDATE 11/14/2013 : Part of the first proposed answer was wrong. The method proposed only works for $n<7$, and I currently have no complete solution for $n\geq 7$. All that is corrected in the updated version below.

Let us put

$$ Q_n(a_1,a_2,\ldots ,a_n)= a^2_{1}+a^2_{2}+\cdots+a^2_{n}-2a_{1}a_{2}-2a_{2}a_{3}-\cdots-2a_{n-2}a_{n-1}-2a_{n-1}a_{n}-2a_na_1 $$


$$ T_n(a_1,a_2,a_3,\ldots,a_n)= Q_n(a_1,a_2,\ldots,a_n)+\frac{(a_1+a_2+a_3+\ldots +a_n)^2}{n} $$

For $3\leq n\leq 6$, the minimum is $-\frac{1}{n}$ (attained when all coordinates are equal to $\frac{1}{n}$), because of

$$ \begin{array}{lcl} T_6(a_1,a_2,\ldots,a_6)&=& \frac{1}{42}\bigg(-5a_1+a_2+a_3+a_4-5a_5+7a_6\bigg)^2 \\ & & +\frac{1}{28}\bigg(-3a_1+2a_2+2a_3-5a_4+4a_5\bigg)^2 +\frac{1}{4}\bigg(-a_1+2a_2-2a_3+a_2\bigg)^2 \\ T_5(a_1,a_2,\ldots,a_5) &=& \frac{1}{30}\bigg(-4a_1+a_2+a_3-4a_4+6a_5\bigg)^2 +\frac{1}{6}\bigg(-a_1+a_2-2a_3+2a_4\bigg)^2 \\ & & +\frac{1}{2}\bigg(-a_2+a_3\bigg)^2+\frac{1}{2}\bigg(-a_1+a_2\bigg)^2 \\ & & \\ T_4(a_1,a_2,a_3,a_4) &=& \frac{1}{20}\bigg(-3a_1+a_2-3a_3+5a_4\bigg)^2 +\frac{1}{20}\bigg(-a_1-3a_2+4a_3\bigg)^2 \\ & & +\frac{3}{4}\bigg(-a_1+a_2\bigg)^2 \\ & & \\ T_3(a_1,a_2,a_3) &=& \frac{1}{3}\bigg(-a_1+a_2+2a_3\bigg)^2 +\bigg(-a_1+a_2\bigg)^2 \\ \end{array} $$

Unfortunately, this method does not work any more for $n \geq 7$. Indeed, in that case the minimum is $\leq -\frac{1}{6}$ (because $Q_n(0,\ldots,0,\frac{1}{6},\frac{1}{3},\frac{1}{3},\frac{1}{6})= -\frac{1}{6}$) ; on the other hand, the polynomial

$$ R_n(a_1,a_2,\ldots ,a_n)=Q_n(a_1,a_2,\ldots,a_n)+\frac{(a_1+a_2+a_3+\ldots +a_n)^2}{6} $$

is not nonnegative on ${\mathbb R}^n$ any more. For example, for $n=7$ we have

$$ R_7(4, 1, 0, -1, 0, 3, 5)=(-2) < 0 $$

  • $\begingroup$ Hello,$a_{2},a_{3}$ coefficient is wrong, and this problem is $$a^2_{1}+a^2_{2}+\cdots+a^2_{n}-2a_{1}a_{2}-2a_{2}a_{3}-\cdots-2a_{n-1}a_{n}-2a_{n}a_{1}$$ is not $$a^2_{1}+a^2_{2}+\cdots+a^2_{n}-2a_{1}a_{2}-2a_{2}a_{3}-\cdots-2a_{n-1}a_{n} $$ $\endgroup$ – math110 Nov 14 '13 at 5:32
  • $\begingroup$ @math110 Indeed, this method works only when $n<7$, I have corrected my answer. $\endgroup$ – Ewan Delanoy Nov 14 '13 at 13:32
  • 1
    $\begingroup$ @EwanDelanoy, for Lagrange multipliers, in this problem $(1/n,1/n,1/n,1/n..)$ is an eigenvector for the Gram matrix of this quadratic form. As a result, that is the only critical point of $Q$ in the plane. For $n \geq 7,$ we know the minimum occurs on the boundary of the set, meaning at least one coordinate is zero, let $a_n = 0.$ Then the minimum is given by your solution to math.stackexchange.com/questions/563804/… $\endgroup$ – Will Jagy Nov 14 '13 at 20:26

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