Suppose $a_1,\dots,a_n$ are positive integers. Trivially one has that $$ \sum_{i=1}^n a_i^2 \leq \left (\sum_{i=1}^n a_i \right)^2 $$ I am wondering whether it is possible to make it somehow sharper, i.e. find some constant $0<C<1$ and $D>0$ such that $$ \sum_{i=1}^n a_i^2 \leq C \left (\sum_{i=1}^n a_i \right)^2 - D. $$ Any comment/idea/counterexample is more than welcome!


No, since when all but one of the $a_i$ are $0$, we have equality.

Edit: You state specifically that the $a_i$ are positive integers. Setting $a_2=\cdots=a_n=1$ gives a counterexample for sufficiently large $a_1$, since $$\left(\sum_{i=1}^n a_i\right)^2=(a_1+n-1)^2=a_1^2+2a_i(n-1)+(n-1)^2$$ and the ratio of this to $$\sum_{i=1}^n a_i=a_1+(n-1)$$ goes to $1$ as $a_1\to\infty$.

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    $\begingroup$ ... but the integers do not contain sufficiently small positive $\epsilon < 1$ $\endgroup$ – martini Dec 10 '13 at 10:04
  • $\begingroup$ @martini Well, it seems I cannot read. I'll fix that. $\endgroup$ – Alex Becker Dec 10 '13 at 10:05
  • $\begingroup$ what about the constant $D$? $\endgroup$ – alezok Dec 10 '13 at 10:58
  • $\begingroup$ @AleZok Even $D=0$ does not work, so certainly $D>0$ does not work. $\endgroup$ – Alex Becker Dec 10 '13 at 11:07

Let $a_1 = k$ and $a_i = 1$ for $i \ge 2$. Then the left hand side equals $$ L := k^2 + (n-1) $$ and the right hand side $$ R := \bigl(k + (n-1)\bigr)^2 = k^2 + 2k(n-1) + (n-1)^2 $$ Now for $k \to \infty$ $$ \frac RL = \frac{k^2 + 2k(n-1) + (n-1)^2}{k^2 + (n-1)} \to 1 $$ That is, there is no such $C$.

For $D$, note that \begin{align*} R &= \left(\sum_i a_i\right)^2\\ &= L + \sum_{i\ne j} a_i a_j\\ &\ge L + \frac{n(n-1)}2 \end{align*} (Note that $a_ia_j \ge 1$ for every pair $i,j$.

  • $\begingroup$ what about the constant $D$? $\endgroup$ – alezok Dec 10 '13 at 10:57
  • $\begingroup$ @AleZok Added something $\endgroup$ – martini Dec 10 '13 at 11:03

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