Inequality problem involving nth harmonic number

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 $\, n \geq 1, \,$

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

Thank you.

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

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.
• @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. – Jack D'Aurizio Apr 26 '17 at 18:48