Does ${E}[\max\{X, Y\}] \geq {E}[\max\{X, Z\}]$ imply $\max\{{E}[X], {E}[Y]\} \geq \max\{{E}[X], {E}[Z]\}$?

Question

Does $$\operatorname{E}[\max\{X, Y\}] \geq \operatorname{E}[\max\{X, Z\}]$$ imply $$\max\{\operatorname{E}[X], \operatorname{E}[Y]\} \geq \max\{\operatorname{E}[X], \operatorname{E}[Z]\}$$ (under the assumption that all the expectations are well defined)?

I don't know whether or not this should be true. My intuition is that if $Y$ and $Z$ have wildly different variations, we could construct a counterexample. If if is true, then I have tried proofs beginning with $\max\{X, Y\} \geq X \implies \operatorname{E}[\max\{X, Y\}] \geq \operatorname{E}[X]$ and similar for the other three cases, but I haven't made any progress worth showing beyond that.

How about the intermediate result that $\operatorname{E}[\max\{X, Y\}] \geq \operatorname{E}[\max\{X, Z\}]$ implies $\operatorname{E}[Y] \geq \operatorname{E}[Z]$?

Answer 1

$X \sim N(0,1), Y=-X, Z=|X|$ gives a counter-example.

Answer 2

The answer is No.

HINT: Let $X$ and $Y$ be random variables that satisfy $P[X=0]=P[Y=0]=\frac{1}{2}$, and $P[X=1]=P[Y=1]=\frac{1}{2}$, and in addition, that also satisfy $X+Y=1$.

Then let $Z$ be a random variable that is $\frac{2}{3}$ everywhere.