How to prove the following inequalities? Thanks for your time. I am interested in various ways/techniques/tricks/methods (induction, convexity, concavity, maximum, minimum, geometry, trigonometry, ...) for proving the  inequalities and their generalizations
(1) $\sqrt{1-x_1^2 -y_1^2}  + \sqrt{1-x_2^2 -y_2^2} + \sqrt{1-x_3^2 -y_3^2}  \le 3\,\sqrt{1-\left(\frac{x_1+x_2+x_3}{3}\right)^2 - \left(\frac{y_1+y_2+y_3}{3}\right)^2} $
I think for $0 \le x_1,y_1,z_1 \le 1$ 
(2) $\sqrt{x_1^2 + y_1^2} + \sqrt{x_2^2 + y_2^2} + \sqrt{x_3^2 + y_3^2} \ge 3 \sqrt{\left(\frac{x_1+x_2+x_3}{3}\right)^2 + \left(\frac{y_1+y_2+y_3}{3}\right)^2}$
for all $x_1,y_1,z_1$
 A: 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.
A: The proof of the second inequality:
Let $r_1=[x_1,y_1]^T, r_2=[x_2,y_2]^T, r_3=[x_3,y_3]^T$ be three vectors in $\Re ^2$. Then the second inequality is equivalent to the following one.
$$\|r_1\|_2+\|r_2\|_2+\|r_3\|_2 \ge \|r_1+r_2+r_3\|_2$$
This inequality is obvious. It is known as the triangle inequality or Cauchy-Schwarz inequality.
A: For (1) I think you need to use the concavity of the function $f(t)=\sqrt{1-t^2}$, i.e. the inequality:
$$ f(\lambda_1 t_1+\lambda_2t_2+\lambda_3t_3) \geq \lambda_1 f(t_1) + \lambda_2 f(t_2)+ \lambda_3 f(t_3) $$
for $t_1,t_2,t_3\in [-1,1]$ and $\lambda_1,\lambda_2,\lambda_3\in [0,1]$ s.t. $\sum_{k=1}^3 \lambda_k=1$; and also the monotonicity of $f(t)$ in $[0,1]$.
