Using the definition of a concave function prove that $f(x)=4-x^2$ is concave (do not use derivative). 
Let $D=[-2,2]$ and $f:D\rightarrow \mathbb{R}$ be $f(x)=4-x^2$. Sketch this function.Using the definition of a concave function prove that it is concave (do not use derivative).

Attempt:
$f(x)=4-x^2$ is a down-facing parabola with origin at $(0,4)$. I know that. But what $D=[-2,2]$ is given for. Is it domain or a point? 
Then, how do I prove that $f(x)$ is concave using the definition of a concave function? I got the inequality which should hold for $f(x)$ to be concave:
For two distinct non-negative values of $x  (u$ and $v$) 
$f(u)=4-u^2$  and $f(v)=4-v^2$
Condition of a concave function:
$ \lambda(4-u^2)+(1-\lambda)(4-v^2)\leq4-[(\lambda u+(1-\lambda)v]^2$
I do not know what to do next.

If you expand your inequality, and fiddle around you can end up with
$$
(\lambda u-\lambda v)^2\leq (\sqrt{\lambda}u-\sqrt{\lambda}v)^2.
$$
Without loss of generality, you may assume that $u\geq v$.  This allows you to drop the squares.  Another manipulation gives you something fairly obvious.  Now, work your steps backwards to give a valid proof.
