请教数学期望问题

It's a special case of Jensen's inequality:
    f(E[X])<=E[f(X)] for convex f.
Since Z=max(X1,X2) is a convex function of X=(X1,X2), then E[max(X1,X2)]>=max(E[X1],E[X2]).
The easiest way to think about it is that you treat the expection as an integral. Since max(x1,x2)>=x1 and max(x1,x2)>=x2, given the joint distribution of X1 and X2, the expectation of max(X1,X2) should be no less than the latter. If equality holds, that means max(X1,X2)=X1 or max(X1,X2)=X2 with probability 1.

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原来中文书的“几乎处处相等”是这么来的...

And in a probabilistic context,
almost everywhere (a.e.) = almost surely (a.s.) = with probability 1.

Is there a premise that X1 and X2 are independent?

No.
If the joint pdf exists, we have
∫max(x1,x2)*f(x1,x2)*dx1*dx2>=∫x1*f(x1,x2)*dx1*dx2.
For a general case, it is
∫max(x1,x2)μ(dx1,dx2)>=∫x1*μ(dx1,dx2).
We never consider the independence between X1 and X2. The joint distribution is enough.

consider two planes z=x1 and z=x2, which intersects at x1=x2=z (a line)
what you need is two half-planes, shaped like a V...

A real-valued function f defined on an convex subset C of some vector space is called convex, if for any two points x and y in its domain C and any t in [0,1], we have
    f(tx+(1-t)y)<=tf(x)+(1-t)f(y).
For x=(x1,x2), y=(y1,y2) and f(x)=max(x1,x2), f(tx+(1-t)y)=max(tx1+(1-t)y1,tx2+(1-t)y2).
On the other hand, tf(x)+(1-t)f(y)=max(tx1,tx2)+max((1-t)y1,(1-t)y2)=max(tx1+(1-t)y1,tx1+(1-t)y2,tx2+(1-t)y1,tx2+(1-t)y2).
Since the two-element set {tx1+(1-t)y1,tx2+(1-t)y2} is a subset of the four-element set{tx1+(1-t)y1,tx1+(1-t)y2,tx2+(1-t)y1,tx2+(1-t)y2}, then the maximum value of the former is no larger than that of the latter, i.e., max(x1,x2) is a convex function.