Show that $$x^{2013}y+xy^{2013} \leqslant x^{2014}+y^{2014}$$
I know that this seems to be just an application of the rearrangement inequality, what I wanted to ask is that what does it actually mean when in this case one could say that "by symmetry" we can assume $x \leqslant y$ or "WLOG" $x \leqslant y$? This always trips me off a bit.
On
Let $P(x,y)$ denote the assertion that the inequality holds for $(x,y)$.
Then for all $x,y$, $P(x,y)\iff P(y,x)$.
So if you prove $P(x,y)$ for all $x,y$ such that $x\leq y$, you also get $P(y,x)$ for those same $x,y$'s.
But now $x,y$ are just names of variables, so you might as well say that we get $P(x,y)$ whenever $y\leq x$. Ha, but now we know $P(x,y)$ holds if $x\leq y$, and it holds if $y\leq x$.
Therefore it holds no matter what.
"To prove that for all $x$, $P(x)$ holds, we may wlog assume that $Q(x)$ holds" in general means that "for all $x$, $P(x)$" is easily deducible from "for all $x$ such that $Q(x)$ holds, $P(x)$ also holds".
Note that this contains some amount of subjectivity ("easily") and so whether "wlog" statements are relevant/valid will depend on who you're talking to.
For instance, imagine you're attempting a proof of Fermat's last theorem "if $n\geq 3$ and $xyz >1$, then $x^n+y^n\neq z^n$". Then to someone who has some amount of experience with number theory, you might say "assume wlog that $n=4$ or is an odd prime".
But of course, an actual proof should be required to show that this is enough : "wlog" just means "This additional proof should be easy for you".
On
To add some more concrete detail for this specific case to the fine given answers, suppose $x\le y$ and
$$x^{2013}y+xy^{2013} \leqslant x^{2014}+y^{2014}$$
now if $y\le x$ by $\bar x=y$ and $\bar y=x$ that is $\bar x\le \bar y$ we obtain
$$\bar y^{2013}\bar x+\bar y\bar x^{2013} \leqslant \bar y^{2014}+\bar x^{2014}$$
that is
$$\bar x^{2013}\bar y+\bar x\bar y^{2013} \leqslant \bar x^{2014}+\bar y^{2014}$$
which is equivalent to the first one.
$x$ and $y$ are just the names of the variables which play a symmetric role here. If you change their names, you get exactly the same exercise (try to write $x$ instead of every $y$ and vice-versa, then arrange). So, in the solution, you can always assume $x$ is the smaller one, because if it is not the case, the entire proof and exercise can be rewritten so that $x$ will be the name of the smaller one.