In a step for proving $\sum\limits_{cyc}{abc}\le5$ when $a + b + c + d + e = 5$. I had the following problem in my book-
$$
abc + bcd + cde + dea + eab \le5
$$
which is to be proved for non - negative real numbers satisfying $a + b + c + d + e = 5$.
Suppose $min(a,b,c,d,e) = e$,
My question is: Is this even possible?
Say on an occasion, $b$ is the least,
It's obvious as the sum is cyclic not symmetric,
\begin{equation}
\begin{split}
f(a,e,c,d,b)
&= aec + ecd + cdb + dba + eab\\
&= abd + abe + ace + bcd + cde\\
\end{split}
\end{equation}
Hence,
$$
f(a,b,c,d,e) \neq f(a,e,c,d,b)
$$
This was the first step to the solution to this problem, in my textbook.
But in my opinion, this is wrong.
May someone explain me where I went wrong, if I did.
Thanks!
 A: I think the proof in book is

Note $$abc + bcd + cde + dea + eab = e(c+a)(b+d)+bc(a+d-e). \quad (*)$$
Suppose $e = \min (a,b,c,d,e)$ and using the AM-GM inequality,
we have $$e(c+a)(b+d)+bc(a+d-e) \leqslant\frac{e(c+a+b+d)^2}{4}+\frac{(b+c+a+d-e)^3}{27} $$
$$=\frac{e(5-e)^2}{4}+\frac{(5-2e)^2}{27}.$$ We need to prove
$$\frac{e(5-e)^2}{4}+\frac{(5-2e)^2}{27} \leqslant 5,$$  equivalent to
$$(e+8)(e-1)^2 \geqslant 0.$$ Done.

Note. Now, what happened if $b$ is the least? The key is identity $(*).$
Indeed, if $b = \min (a,b,c,d,e)$ we write $(*)$ as
$$abc + bcd + cde + dea + eab = b(c+e)(a+d)+de(c+a-b).$$
$$ \leqslant \frac{b(c+e+a+d)^2}{4}+\frac{(d+e+c+a-b)^2}{27}.$$
$$ = \frac{b(5-b)^2}{4}+\frac{(5-2b)^2}{27}.$$
If $c = \min (a,b,c,d,e)?$ We write $(*)$ as
$$abc + bcd + cde + dea + eab=c(b+e)(a+d)+ae(b+d-c).$$
$$\leqslant  \frac{c(5-c)^2}{4}+\frac{(5-2c)^2}{27}.$$
Similar to $c, d,a$ is the least, and using the AM-GM inequality as above.
Conclude. Without loss of generality we can suppose $e = \min(a,b,c,d,e).$
A: While you already have a good answer which shows why it doesn't matter in the specific problem, please note that this is in fact general.
It is true that in ANY cyclic case, we may WLOG take any of the variables to be the minimum (or maximum).  This is because you can cycle through $(a, b, c, d, e)  \rightarrow (b, c, d, e, a) \rightarrow (c, d, e, a, b)  \rightarrow (d, e, a, b, c) \rightarrow (e, a, b, c, d)...$ to get the minimum (or maximum) to occupy the first position or the last position, or any of the positions you desire.
In your particular e.g. what matters is that $f(c, d, e, a, b) = f(a, b, c, d, e)$, so it doesn't matter if $b$ is the smallest instead of $e$.
In fact you can even say WLOG things like let $a, b, c$ be s.t. $b \geqslant \max(a, c)$. Again the logic is the same, in a (cyclic) sequence, you will somewhere always find one number at least as high as its neighbours, and then you can cyclically push that triplet to the front. Or things like WLOG let $b-a$ be the highest among consecutive differences.  If that helps solve the particular inequality in question, great!
