prove that $\int_0^\frac{\pi}{4} \frac{1-\cos x}{x} \,dx < \frac{\pi^2}{64}$ prove that $\int_0^\frac{\pi}{4} \frac{1-\cos x}{x} \,dx <  \frac{\pi^2}{64}$
I showed that in 
$$ \forall x \in [0,\frac{\pi}{4}] \quad  \int_0^\frac{\pi}{4} \frac{1-\cos x}{x} \,dx \le \int_0^\frac{\pi}{4} \frac{1-\cos(\frac{\pi}{4})}{\frac{\pi}{4}} \,dx = 1 -\frac{\sqrt2}{2} $$
but it's not good enough

Since:
$$\frac{\sin x}{x}=\prod_{n\geq 1}\left(1-\frac{x^2}{n^2\pi^2}\right) $$
over the interval $\left[0,\frac{\pi}{4}\right]$ we have:
$$ \frac{1-\cos x}{x}=\left(\frac{\sin(x/2)}{x/2}\right)^2\cdot\frac{x}{2}\leq\left(1-\frac{x^2}{4\pi^2}\right)^2\frac{x}{2} $$
so:
$$ \int_{0}^{\frac{\pi}{4}}\frac{1-\cos x}{x}\,dx \color{red}{\leq} \int_{0}^{\pi/4}\left(1-\frac{x^2}{4\pi^2}\right)^2\frac{x}{2}\,dx = \frac{\pi^2}{64}-\frac{191\,\pi^2}{3\cdot 2^{18}}.$$
