Consider an arbitrary non-negative random variable $Z \in L^\infty$ such that $E[Z]=1$ and $\lVert Z \rVert_{2}=2$.
Does there exist a non-negative random variable $Y \in L^\infty$ such that $E[Y]=1$ and $\lVert \lambda Y + (1-\lambda)Z \rVert_{2} > 2$ for all $\lambda \in (0,1)$?
My thinking is yes, because the $L^2$-norm of $Y$ can be as large as we like. But I am finding it difficult to prove. My idea is that since $Z$ is in $L^\infty$, you can construct $Y$ so that: $Y>Z$ for large values of $Z$, $Y<Z$ for small values of $Z$, and otherwise $Y=Z$ (making sure $\mathbb{E}[Y]=1$). But writing this down formally is tricky. Is this the right way to go about it? Perhaps there is a way of proceeding via contradiction.
Extension: What happens if $Z \in L^2$ (and so not necessarily bounded)?
The answer is no, e.g., $\Omega=\{\omega_1, \omega_2\}$ and $\mathbb{P}[\{\omega_1\}] = \frac{1}{2}$. Consider $Z=2\mathbf{1}_{\{\omega=\omega_1\}}$.