What is the convex-hull of the set $\{ (n,\varphi(n)) : n\in \mathbb N \} \subset \mathbb R^2$ I know that set
$$
E=\{ (n,\varphi(n)) : n\in \mathbb N \} \subset \mathbb R^2
$$
has infinitely many points on the line $y=x-1$, which suggests this line to be included in the upper part of the convex-hull. However I don't really see what's going on at the bottom. That is what is the convex-hull of $E$?
Thanks.
 A: Let $0 < c < 1$. Then there are infinitely many $n$ such that $\varphi(n) < cn$, and thus the line $y = c(x-1) + 1$ $(x \geq 1$) is included in the convex hull. 
The convex hull is hence the triangle-shaped area determined by the lines $y = x$, ($x \geq 1$) and $y = 1$.
EDIT: Proof for the claim above. If $P_k = p_1 p_2 \ldots p_k$ is the product of first $k$ primes, then $\varphi(P_k) / P_k = \prod_{i = 1}^k (1 - \frac{1}{p_i})$ converges to zero as $k \rightarrow \infty$. See for example here.
A: Relevant facts:
For infinitely many $n$, we have
$$
\varphi(n) < \frac{n}{e^\gamma \log \log n} \tag{1}
$$
See here.
But the constant $e^\gamma$ doesn't really matter.
In fact, the simpler condition given in spin's answer suffices; i.e. that for any $0 < c < 1$, there are infinitely many $n$ such that
$$
\varphi(n) < cn \tag{2}
$$
Claim:
The convex hull $\text{conv } E$ is given by
$$
\{(1,1)\} \cup \left\{ (x,y) \; : \; 1 < y < x. \right\}.
$$
Proof.
Fix any real $x > 1$.  For any $n$, $\text{conv } E$ contains
the line segment between $(1,1)$ and $(n, \varphi(n))$.
In particular, it contains the point
$\left(x, \frac{\varphi(n) - 1}{n-1} (x-1) + 1\right)$.
For any $0 < c < 1$, find an $n$ satisfying (2).  Then $\text{conv } E$ contains the point $(x,y)$ with
$$
y = \frac{\varphi(n) - 1}{n-1} (x-1) + 1
<
\frac{cn - 1}{n-1} (x-1) + 1
<
\frac{cn}{n / 2} (x-1) + 1
= \frac{c}{2} (x - 1) + 1
$$
As $c \to 0$, this gives us points arbitrarily close to $(x, 1)$ in $\text{conv } E$.
On the other hand, if we pick $n$ to be a large prime,
$$
y = \frac{\varphi(n) - 1}{n-1} (x-1) + 1
=
\frac{n - 2}{n-1} (x-1) + 1
=
\left(1 - \frac{1}{n-1} \right) (x-1) + 1
$$
Since there are infinitely many primes, this gives us points arbitrarily close to $(x, x)$ in $\text{conv } E$.
For any two points $(x, y_1)$ and $(x,y_2)$, the segment between them is contained in $\text{conv } E$.  Therefore,
$\{x\} \times (1, x) \subset \text{conv } E$.
$x$ was arbitrary, so we have $\{ (x,y) \; : \: 1 < y < x \} \subset \text{conv } E$.
Adding the point $(1,1)$, we arrive at a convex set containing $(n, \phi(n))$ for all $n$, so this must be the entire convex hull.
A: Here is a more precise bound in case you find it useful, even though the question is already answered.
It was shown by Nicolas, that for infinitely many positive integers $n$ we have
$\displaystyle \frac{n}{\varphi(n)}>e^{\gamma}\log \log n$ where $\gamma$ is the Euler constant. You can find details in 
J.-L. Nicolas. Petites valeurs de la fonction d’Euler. J. Number Theory 17 (1983), 375–388.
