**It is possible to prove that the inequality holds for sufficiently large $n$ with $1/4$ replaced by $1/4+\epsilon$ for any $\epsilon>0$ .**

**(Update - see below for a stronger result conditional on the irrationality measure of $\pi$)**

In fact we have:

**Theorem**

For any $\epsilon>0$ and large enough $n$, depending on $\epsilon$ only, the following inequality holds for any real $a_i\geq 0$

$$|\sum_{k=1}^n a_k e^{ik}|\leq (1/2+\epsilon) \sqrt n (\sum_{k=1}^{n}a_k^2)^{1/2}.$$

**Proof**

We know that $$|\sum_{k=1}^n a_k e^{ik}| = e^{ i\theta}\sum_{k=1}^n a_k e^{ik}$$ for some $\theta \in [0,2\pi]$. Hence

$$|\sum_{k=1}^n a_k e^{ik}| = \sum_{k=1}^n a_k e^{i(k+\theta)}=\sum_{k=1}^n a_k\cos(k+\theta).$$

Define a function $ \cos_{+}:\mathbb{R}\to[0,1]$ by $\cos_{+}(x)=\cos(x)$ if $\cos(x)\geq 0$ and $0$ otherwise.

Then using $a_k\geq0$ and Cauchy's Theorem,

$$\sum_{k=1}^n a_k\cos(k+\theta)\leq\sum_{k=1}^n a_k\cos_+(k+\theta)\leq(\sum_{k=1}^n a_k^2)^{1/2}(\sum_{k=1}^n {\cos_+}^2(k+\theta))^{1/2}.$$

Now the equidistribution theorem says that $\{\frac{i}{2\pi}\}$ is uniformly distributed in $[0,1]$. Hence by the Riemann integral criterion for equidistribution,

$$\lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^{n} f\left(s_k\right) = \frac{1}{b-a}\int_a^b f(x)\,dx$$

with $f(x)={\cos_+}^2(2\pi x+\theta)$, $s_k=\{\frac{k}{2\pi}\}$ and $a=0$, $b=1$.

In other words,

\begin{equation}
\begin{split}
& \lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^n {\cos_+}^2(k+\theta)=\int_0^1 {\cos_+}^2(2\pi x+\theta)\,dx =\frac{1}{2\pi}\int_{\theta}^{2\pi+\theta} {\cos_+}^2(\phi)\,d\phi \\ & =\frac{1}{2\pi}\int_{-\pi/2}^{+\pi/2} \cos^2(\phi)\,d\phi=1/4.
\end{split}
\end{equation}

Hence

$$\frac{|\sum_{k=1}^n a_k e^{ik}|}{\sqrt n (\sum_{k=1}^n a_k^2)^{1/2} }\leq \left(\frac{1}{n}\sum_{k=1}^n {\cos_+}^2(k+\theta)\right)^{1/2}\rightarrow 1/2 \ \text{ as } \ n\rightarrow \infty$$

and the result follows.

$\blacksquare$

**In addition I can prove the following estimate, conditional on the irrationality measure of $\pi$ being less than 3:**

**Theorem**

If the irrationality measure for $\pi$, $\mu(\pi)$ is strictly less than 3 then $$\frac{|\sum_{k=1}^n a_k e^{ik}|^2}{ \sum_{k=1}^n a_k^2 }\leq\frac{n}{4}+D.$$

for some fixed constant $D$.

**Proof**

We know from the above that

$$|\sum_{k=1}^n a_k e^{ik}|^2\leq (\sum_{k=1}^n a_k^2) \sum_{k=1}^n {\cos_+}^2(k+\theta)$$ for some $\theta \in [0,2\pi]$.

Also, ${\cos_+}^2(x) = (\cos^2 x+\cos x|\cos x|)/2$.

Hence $$\sum_{k=1}^n {\cos_+}^2(k+\theta)=\sum_{k=1}^n (\cos^2 (k+\theta)+\cos (k+\theta)|\cos (k+\theta)|)/2\\=\frac{1}{2}\sum_{k=1}^n \cos^2 (k+\theta)+\frac{1}{2}\sum_{k=1}^n (\cos (k+\theta)|\cos (k+\theta)|)$$.

Clearly $\sum_{k=1}^n \cos^2 (k+\theta)\leq n/2+B$ for some constant $B$ so it remains to bound the other term which is more complicated.

Let $f(x) = |\cos{(x)}| \cos{(x)}$. Here it is noted by Andreas that the expression $|\sin{(x)}| \sin{(x)}$ can be written as a Fourier series, which we modify to provide a series for $f$,

$$
f(x) = \frac{8}{\pi}\sum_{m=0}^\infty \frac{(-1)^m}{4(2m+1)-(2m+1)^3} \cos((2m+1)x).
$$
Now we can sum
$$
S_n = \sum_{k=1}^{n} f{(k + \phi)} = \frac{8}{\pi}\sum_{m=0}^\infty \frac{(-1)^n}{4(2m+1)-(2m+1)^3} \sum_{k=1}^{n}\cos((2m+1)(k + \phi))
$$
where
$$
\sum_{k=1}^{n}\cos((2m+1)(k + \phi)) =
\frac{\sin(n(m + \frac12)) \cdot \cos((1 + n + 2 \phi)(m + \frac12)) }{ \sin(m + \frac12)}.
$$

Thus

$$
S_n = \frac{8}{\pi}\sum_{m=0}^\infty \frac{(-1)^n}{4(2m+1)-(2m+1)^3} \frac{\sin(n(m + \frac12)) \cdot \cos((1 + n + 2 \phi)(m + \frac12)) }{ \sin(m + \frac12)}.
$$
Taking absolute values and using the triangle inequality
$$
|S_n| \leq \frac{8}{\pi}\sum_{m=0}^\infty \frac{1}{|4(2m+1)-(2m+1)^3|\sin(m + \frac12)|}\\=\frac{8}{\pi}\sum_{m=0}^\infty \frac{|2\cos (m+\frac12)|}{|4(2m+1)-(2m+1)^3||\sin(2 m + 1)|}
\\ \leq \frac{16}{\pi}\sum_{m=0}^\infty \frac{1}{|(2m+1)^3-4(2m+1)||\sin(2 m + 1)|}
\\ =\frac{16}{3 \pi \sin 1}+\frac{16}{15 \pi \sin 3}+\frac{16}{\pi}\sum_{m=2}^\infty \frac{1}{|(2m+1)^3-4(2m+1)||\sin(2 m + 1)|}
\\ \leq \frac{16}{3 \pi \sin 1}+\frac{16}{15 \pi \sin 3}+\frac{16}{\pi}\sum_{m=2}^\infty \frac{1.3}{(2m+1)^3|\sin(2 m + 1)|}
$$

To complete the estimate we note that Theorem 5 of Max A. Alekseyev's paper "On convergence of the Flint Hills series" implies that if $\mu(\pi)<3$ then $\sum_{n=1}^{\infty}\frac{1}{n^3|\sin n|}$ converges, hence $\sum_{n=1}^{\infty}\frac{1}{(2n+1)^3|\sin (2n+1)|}$ converges also and we have

$$|S_n| \leq C$$ for some fixed constant $C$. Combining the two estimates we have on the assumption that $\mu(\pi)<3$,

$$\sum_{k=1}^n {\cos_+}^2(k+\theta)=\frac{1}{2}\sum_{k=1}^n \cos^2 (k+\theta)+\frac{1}{2}\sum_{k=1}^n (\cos (k+\theta)|\cos (k+\theta)|)\\ \leq n/4+(B+C)/2$$

and the result is proved. $\blacksquare$

Unfortunately although most irrational numbers have irrationality measure 2, and this is probably the true value of $\mu(\pi)$, the best upper bound for $\mu(\pi)$ is 7.103205334137 due to Doron Zeilberger and Wadim Zudilin - see here for their paper so we are a long way from being able to prove the inequality this way at least.