Given $X$ is a continuous random variable taking values in the reals, and $\phi(t)$ is its characteristic fiction, that is, $\phi(t):=\Bbb E(\exp(iXt))$, then prove that $$\Re (1-\phi(t))\ge \frac14\Re(1-\phi(2t)).$$
Reducing everything, I get what is to be shown is actually $$\Bbb E|\cos tY|\le \Bbb E|\cos tY|^2+\frac14.$$ I don't know how to get over that $\dfrac14$. Is there any inequality that can yield an additive constant on one side? As far as I know, neither Chebyshev nor Hölder can.
It's just a trigonometric inequality, obtained from the double-angle formulae. On the left hand side, we use
$$1 - \cos \varphi = 2\sin^2 \frac{\varphi}{2},$$
and on the right hand side,
\begin{align} \frac{1-\cos (2\varphi)}{4} &= \frac{\sin^2 \varphi}{2} \\ &= \frac{\bigl(2\sin \frac{\varphi}{2}\cos \frac{\varphi}{2}\bigr)^2}{2} \\ &= 2\sin^2 \frac{\varphi}{2} \cos^2 \frac{\varphi}{2}. \end{align}
Since $\cos^2 \frac{\varphi}{2} \leqslant 1$, we have a pointwise inequality for the integands, and integrating with respect to a positive measure, in particular a probability measure, preserves the pointwise inequality.