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.

$\endgroup$
  • 1
    $\begingroup$ Try the Cauchy-Schwarz on a modified sequence; eg: $a_j-\dfrac{1}{2\sqrt{j}}$. $\endgroup$ – Aravind Apr 26 '17 at 16:27
  • 1
    $\begingroup$ Can you please post your solution? $\endgroup$ – rationalpi Apr 26 '17 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 '17 at 18:01
2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – rationalpi Apr 26 '17 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 '17 at 18:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.