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 ≤ a^{3} + b^{3} + c^{3}. First, plug in w = (xyz)^{1/3} in part (a). One obtains
4xyz(xyz)^{1/3} ≤ x^{4} + y^{4} + z^{4} + (xyz)^{4/3}. Notice (xyz)^{1/3} always exists, even if xyz is negative.

Rearranging terms, one obtains 4(xyz)^{4/3} ≤ x^{4} + y^{4} + z^{4} + (xyz)^{4/3}, i.e.
3(xyz)^{4/3} ≤ x^{4} + y^{4} + z^{4}

Apply this last inequality with x = a^{3/4}, y = b^{3/4}, z= c^{3/4} to obtain what you want.

I believe this problem is harder than what you may expect in an exam.