Question: I know this is a pretty simple question, but I'm just not getting the textbook... I'm taking a basic CS course and on one of the problems (not an assigned homework problem, just one I'm practicing on), it says that 

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> on the set of all integers, the relation $x \ne$ y is symmetric but not transitive where $(x,y) \in \Bbb R$. 
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I understand why it's symmetric, but why is it not transitive? I'm thinking it has something to do with the definition of transitive including ALL $a,b,c \in A$, but I'm not sure. I understand transitive in the context of a finite set, but something about applying it to all integers is throwing me off.

Answer: 
Let $f(x)=x^p$
By convexity $$f(\frac{X+Y}{2})\leq \frac{1}{2} f(X)+\frac{1}{2} f(Y) \implies \frac{(X+Y)^p}{2^p}\leq\frac{1}{2} X^p+\frac{1}{2}Y^p$$
The result follows.
If $p>1$, then by Holder inequality,
\begin{align\*}
X+Y &\le (X^p+Y^p)^{\frac{1}{p}} 2^{1-\frac{1}{p}}.
\end{align\*}
That is,
\begin{align\*}
(X+Y)^p \le 2^{p-1} (X^p+Y^p).
\end{align\*}
For $0 \le p \le 1$, we note that
\begin{align\*}
\left(\frac{X}{X+Y} \right)^p + \left(\frac{Y}{X+Y} \right)^p \ge \frac{X}{X+Y}+ \frac{Y}{X+Y}=1,
\end{align\*}
and then
\begin{align\*}
(X+Y)^p \le X^p+Y^p.
\end{align\*}