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.

To be concave $f(x)=4-x^2$ should satisfy the condition for a concave function:
For two distinct values of x (u and v) such that
$f(u)=4-u^2$ and $f(v)=4-v^2$ the following inequality should be true:
$$\lambda f(u)+(1-\lambda)f(v)\leq f(\lambda u+(1-\lambda)v), \;\text{for}\; 0<\lambda<1$$  
which turns into the inequality below:
$$
\lambda(4-u^2)+(1-\lambda)(4-v^2)\leq4-[(\lambda u+(1-\lambda)v]^2
$$
To show that the above inequality is true, first, I expanded it and made it look like the expression Joe posted(remember that $0<\lambda<1$ by definition of a concave function). After expanding LHS I got:
$4\lambda -\lambda u^2 +4-4\lambda -v^2+\lambda v^2\leq4-[(\lambda u+(1-\lambda)v]^2$
Then, I canceled out the terms $4\lambda , 4$ and subtracted the RHS from the LHS:
$-\lambda u^2-v^2+\lambda v^2+(\lambda+v-\lambda v)^2\leq0$
After expanding the expression in parenthesis the following terms cancel out: $v^2,\lambda v^2$. Rearranging the remaining terms I got:
$\lambda^2 u^2-2\lambda^2 uv+\lambda^2 v^2\leq\lambda u^2-2\lambda uv+\lambda v^2$
Which turns into the expression Joe posted:
$(\lambda u-\lambda v)^2\leq (\sqrt{\lambda}u-\sqrt{\lambda}v)^2$
Then, I factored out $\lambda 's$ and subtracted the RHS from the LHS:
$\lambda ^2(u-v)^2-\lambda (u-v)^2\leq0$
Finally I factored out $\lambda (u-v)^2$ and got:
$\lambda (u-v)^2(\lambda - 1)\leq0$
which is definitely true because by definition $0<\lambda<1$ which makes the LHS strictly negative:
$\lambda (u-v)^2(\lambda-1)<0$
I fiddled around with the terms of the inequality that should be proved for the $f(x)=4-x^2$ to be concave and turned it into the form that clearly shows that it is true for $0<\lambda<1$ which is a part of the condition for a concave function. 
