I gave yet another explanation on why the negation of p implies q is p and not q, and is not p implies not q. This used the statement "If I am elected, I will free all the prisonners" as a starting point.
I discussed Problem 1.51 and 1.31 from the book. Embarassingly, I couldn't finish 1.31 b) . Here is one way:
Let a,b,c be reals. I prove that 3abc ≤ a3 + b3 + c3. First, plug in w = (xyz)1/3 in part (a). One obtains 4xyz(xyz)1/3 ≤ x4 + y4 + z4 + (xyz)4/3. Notice (xyz)1/3 always exists, even if xyz is negative.
Rearranging terms, one obtains 4(xyz)4/3 ≤ x4 + y4 + z4 + (xyz)4/3, i.e. 3(xyz)4/3 ≤ x4 + y4 + z4
Apply this last inequality with x = a3/4, y = b3/4, z= c3/4 to obtain what you want.
I believe this problem is harder than what you may expect in an exam.