As mentioned by others, (1) and (2) are convexity/concavity inequalities, see this for example. A generalization of both inequalities which can help to see what is going on is as follows.
For every $z$ in $\mathbb{R}^n$, let $\|z\|$ denote its Euclidean norm.
Consider $x_1$, $x_2$, ..., $x_K$ in $\mathbb{R}^n$, and introduce their barycenter
$$
x=\frac1K\sum_{k=1}^Kx_k.
$$
One sees that (1) is a special case of the fact that, if $\|x_k\|\le1$ for every $k$,
then
$$
\frac1K\sum_{k=1}^K\sqrt{1-\|x_k\|^2}\le \sqrt{1-\|x\|^2}.
$$
Likewise, (2) is a special case of the fact that
$$
\frac1K\sum_{k=1}^K\|x_k\|\ge \|x\|.
$$
The OP asked about the case $n=2$ and $K=3$.
This general formulation indicates that these inequalities are true if (and only if) some specific functions $\varphi$ and $\psi$ defined on $\mathbb{R}^n$ are concave, respectively convex. Here,
$$
\varphi(z)=\sqrt{1-\|z\|^2},
\quad
\psi(z)=\|z\|.
$$
A simple way to prove that a function is concave/convex is to write it as an infimum/supremum of affine functions. Recall that an affine function on $\mathbb{R}^n$ is defined by $z\mapsto\langle a,z\rangle+b$ for a given $a$ in $\mathbb{R}^n$ and a given $b$ in $\mathbb{R}$. Now,
$$
\varphi(z)=\inf_{u\in\mathbb{R}^n}\langle u,z\rangle+\sqrt{1+\|u\|^2},\quad
\psi(z)=\sup_{\|u\|=1}\langle u,z\rangle,
$$
hence the generalizations of (1) and (2) hold.