You are mistaking the inequality. Tao is saying that there is a constant $C > 0$ with $$|\frac{1}{N}\sum_{n \le N} a_n| \le C(\frac{1}{H}\sum_{h \le H} |\frac{1}{N}\sum_{n \le N} a_{n+h}\overline{a_n}|)^{1/2} + C\frac{H}{N}.$$
Actually, I think there should be a $C\frac{1}{H^{1/2}}$ on the right hand side as well (see the link I gave).
Based on where you got to, and using $\sqrt{x+y} \le \sqrt{x}+\sqrt{y}$, it suffices to show $$\frac{1}{N}\sum_{n \le N} |\frac{1}{H}\sum_{h \le H} a_{n+h}|^2 \le \frac{2}{H}\sum_{h \le H} |\frac{1}{N}\sum_{n \le N} a_{n+h}\overline{a_n}|+O(\frac{1}{H})+O(\frac{H}{N}).$$
Merely by expanding the square, the left hand side is $$\frac{1}{H^2}\sum_{h_1,h_2 \le H} \frac{1}{N}\sum_{n \le N} a_{n+h_1}\overline{a_{n+h_2}}.$$ The term $h_1 = h_2$ has contribution $$\frac{1}{H^2}\sum_{h \le H} \frac{1}{N}\sum_{n \le N} |a_{n+h}|^2 = O(\frac{1}{H}),$$ so we can ignore it. We are left with
\begin{align*}
\frac{2}{H^2}\sum_{1 \le h_1 < h_2 \le H} \frac{1}{N}\sum_{n \le N} a_{n+h_1}\overline{a_{n+h_2}} &= \frac{2}{H^2}\sum_{1 \le h_1 < h_2 \le H} \frac{1}{N}\sum_{n=h_1+1}^{N+h_1} a_n\overline{a_{n+h_2-h_1}} \\ &= \frac{2}{H^2}\sum_{1 \le h_1 < h_2 \le H} \left[\frac{1}{N}\sum_{n=1}^N a_n\overline{a_{n+h_2-h_1}} + O(\frac{H}{N})\right].
\end{align*}
The $O(\frac{H}{N})$ term merely comes out of the triple sum (since everything is averaged), so we can ignore it. We are left with $$\frac{2}{H^2} \sum_{1 \le h_1 < h_2 \le H} \frac{1}{N}\sum_{n=1}^N a_n\overline{a_{n+h_2-h_1}} = \frac{2}{H^2}\sum_{h \le H}\sum_{\substack{1 \le h_1 < h_2 \le H \\ h_2-h_2 = h}} \frac{1}{N}\sum_{n=1}^N a_n\overline{a_{n+h}}.$$
Now, we can pull out the sum in $n$ and note that the number of $1 \le h_1 < h_2 \le H$ with $h_2-h_1 = h$ is $H-h$ to get $$\frac{2}{H^2}\sum_{h \le H} (H-h)\frac{1}{N}\sum_{n=1}^N a_n\overline{a_{n+h}},$$ which is upper bounded by $$\frac{2}{H^2}\sum_{h \le H} (H-h)\left|\frac{1}{N}\sum_{n=1}^N a_n\overline{a_{n+h}}\right|,$$ which is of course upper bounded by $$\frac{2}{H}\sum_{h \le H} \left|\frac{1}{N}\sum_{n=1}^N a_n\overline{a_{n+h}}\right|,$$ as desired.
In regards to what this inequality is trying to capture. First, note that it holds in any Hilbert space, i.e. $||\frac{1}{N}\sum_{n \le N} v_n|| \le C(\frac{1}{H}\sum_{h \le H} ||\frac{1}{N}\sum_{n \le N} \langle v_{n+h},v_n \rangle||)^{1/2}+C\frac{1}{H^{1/2}}+C\frac{H}{N}$ (just go through the proof). Now suppose the $v_n$'s are mutually orthogonal. Then the left hand side is $\frac{1}{\sqrt{N}}$, while the right hand side is $O(\frac{1}{H^{1/2}})+O(\frac{H}{N})$ (which actually shows the need for the $O(\frac{1}{H^{1/2}})$ term), where the $O(\frac{1}{H^{1/2}})$ term came from $\frac{1}{N}\sum_{n \le N} \langle v_n,v_n \rangle$, the "diagonal term" in the proof above. The point is that the Van der Corput inequality allows one to get good bounds on $||\frac{1}{N}\sum_{n \le N} v_n||$ when the $v_n$'s are "almost orthogonal" (the preceding sentence shows it specializes to actual orthogonality).
Best Answer
The Polya's proof in my writing.
Let $c_k=\frac{(k+1)^k}{k^{k-1}}.$
Thus, $c_1c_2...c_k=(k+1)^k$ and by AM-GM we obtain: $$\sum_{k=1}^n\sqrt[k]{u_1 u_2 ... u_k}=\sum_{k=1}^n\frac{\sqrt[k]{u_1c_1 u_2c_2 ... u_kc_k}}{k+1}\leq\sum_{k=1}^n\frac{u_1c_1+u_2c_2+...+u_kc_k}{k(k+1)}=$$ $$=\sum_{k=1}^n u_kc_k\sum_{i=k}^n\frac{1}{i(i+1)}=\sum_{k=1}^n u_kc_k\left(\frac{1}{k}-\frac{1}{k+1}+\frac{1}{k+1}-\frac{1}{k+2}+...+\frac{1}{n}-\frac{1}{n+1}\right)<$$ $$<\sum_{k=1}^nu_kc_k\cdot\frac{1}{k}=\sum_{k=1}^nu_k\frac{(k+1)^k}{k^{k-1}}\cdot\frac{1}{k}=\sum_{k=1}^nu_k\left(1+\frac{1}{k}\right)^k<\left(1+\frac{1}{n}\right)^n\sum_{k=1}^nu_k<e\sum_{k=1}^nu_k.$$