Find smallest $c \leq 1$ such that $\operatorname{var} \left(\max(\mathbb{E}[X], X) \right) \leq c \,\operatorname{var}(X)$

Question

Consider random variable $X \geq 0$. Find smallest $c \leq 1$ such that $\operatorname{var} \left(\max(\mathbb{E}[X], X) \right) \leq c \operatorname{var}(X)$.

Intuitively, it seems to be true for some $c < 1$, but I could only prove it for $c = 1$. Any idea?

Answer 1

Any $c <1$ will not be generally correct as an upper bound

Suppose $X = 1$ with probability $p > 0$ and $X=-\frac{p}{1-p}$ with probability $1-p$

then $\mathbb E[X] = 0$ and $\operatorname{var}(X)=\frac{p}{1-p}$

while $\mathbb E[\max(\mathbb{E}[X], X)] = p$ and $\operatorname{var}(\max(\mathbb{E}[X], X)) = p(1-p)$

so $\frac{\operatorname{var}(\max(\mathbb{E}[X], X)) }{\operatorname{var}(X)} = (1-p)^2 > c$ when $0 < p < 1-\sqrt{c}$