4
$\begingroup$

Let $\, a_1, a_2, a_3, \ldots \,$ be a sequence of positive real numbers satisfying $\, \sum_{j = 1}^n a_j \geq \sqrt {n} \,$ for all $\, n \geq 1$. Prove that, for all $\, n \geq 1, \,$

$$\sum_{j = 1}^n a_j^2 > \frac {1}{4} \left( 1 + \frac {1}{2} + \cdots + \frac {1}{n} \right).$$

I tried Cauchy-Schwarz but cant proceed anywhere. Please help!

Thank you.

$\endgroup$
3
  • 1
    $\begingroup$ Try the Cauchy-Schwarz on a modified sequence; eg: $a_j-\dfrac{1}{2\sqrt{j}}$. $\endgroup$
    – Aravind
    Apr 26, 2017 at 16:27
  • 1
    $\begingroup$ Can you please post your solution? $\endgroup$
    – user428700
    Apr 26, 2017 at 17:24
  • 1
    $\begingroup$ @dovakin123: I have posted a solution that proves the claim by summation by parts, avoiding Cauchy-Schwarz. $\endgroup$
    – Jack D'Aurizio
    Apr 26, 2017 at 18:01

1 Answer 1

Reset to default
1
$\begingroup$

Let $\Delta_k=\sqrt{k}-\sqrt{k-1}$ for any $k\in\{1,2,\ldots,n\}$.
We have $\sum_{k=1}^{m}(a_k-\Delta_k)\geq0$. Let $U_m=\sum_{k=1}^{m}(a_k-\Delta_k)$ for any $m\in\{1,2,\ldots,n\}$.
By $a_k = (a_k-\Delta_k)+\Delta_k$ and summation by parts we have:

$$\sum_{k=1}^{n}a_k^2 \geq \sum_{k=1}^{n}\left[\Delta_k^2+2\Delta_k(a_k-\Delta_k)\right]\geq\sum_{k=1}^{n}\Delta_k^2+2\sum_{k=1}^{n-1}U_k\left(\Delta_{k}-\Delta_{k+1}\right)$$ and the RHS is $\geq \sum_{k=1}^{n}\Delta_k^2$ since $U_k\geq 0$ and $\Delta_k$ is decreasing. On the other hand $$ \Delta_k = \frac{1}{\sqrt{k}+\sqrt{k-1}}>\frac{1}{2\sqrt{k}} $$ and the claim readily follows.

$\endgroup$
2
  • $\begingroup$ Thanks a lot, I didn't thought of using Abel Summation can I ask what led you to guess the other sequence? I just wish to hear your motivation for this solution if you like. $\endgroup$
    – user428700
    Apr 26, 2017 at 18:40
  • 1
    $\begingroup$ @dovakin123: I started by trying to apply Aravind's suggestion, but I did not see a straightforward way to conclude by Cauchy-Schwarz. So I modified Aravind's $\frac{1}{2\sqrt{j}}$ into the telescopic term $\sqrt{j}-\sqrt{j-1}$ (very similar) and tried to see if Abel's trick produced something interesting. Indeed, it did. $\endgroup$
    – Jack D'Aurizio
    Apr 26, 2017 at 18:48

You must log in to answer this question.