Concavity and Convexity A set $X \subseteq \mathbb{R}^n$ is said to be convex if $tx + (1-t)y \in X$ for all $x,y \in X$ and $t \in (0,1)$. Given a convex set $X \subseteq \mathbb{R}^n$, a function $f: X  \to \mathbb R$ is said to be concave if $f(tx + (1-t)y) \ge tf(x) + (1-t)f(y)$ for all $x,y \in X$ and $t \in (0,1)$.
1) Show that $f: \mathbb{R}^n \to \mathbb R$ is concave iff 
    $\sum_{i=1}^r t_if(x_i) \le f\left(\sum_{i=1}^r t_ix_i\right)$
for every positive integer r, for all $x_1, \dots, x_r \in \mathbb R^n$, and all $t_1,\dots,t_r \in (0,1)$ with $\sum_{i=1}^r t_i = 1$
2) Use (1) to show that $\prod_{i=1}^r x_i^{t_i} \le \sum_{i=1}^r t_ix_i$
   for all non negative $x_1, \dots, x_r \in \mathbb R$ and all $t_1,\dots,t_r \in (0,1)$ with $\sum_{i=1}^r t_i = 1$
3) Show that the function $f: \mathbb{R}^n \to \mathbb{R}$ is convex iff the set
     $\{(x,r) \in  \mathbb{R}^n \times \mathbb{R}  \mid  f(x) \le r \}$ 
   is convex.
 A: For part 1), try induction on $r$. For 2), think about log. For 3) think about the pairs $(x, f(x)), (y, f(y))$.
I'm being somewhat mysterious because I think these are healthy exercises to solve for oneself.
EDIT:
Hey sorry for the delay. For the first part, the second condition implies concavity so we only need to prove the first condition implies the second. For the base case, this is easy. Suppose it holds for k and pick $x_1, ..., x_k, x_{k+1} \in \mathbb{R}^n $ and $t_1, ..., t_{k+1} \in (0, 1)$ such that $\sum\limits_{i=1}^{k+1} t_i=1$. Now, let $x'_k=\frac{t_kx_k+t_{k+1}x_{k+1}}{t_k+t_{k+1}}$ and $t'_k= t_k+t_{k+1}$. Then $\sum\limits_{i=1}^{k-1} t_i + t'_k=1$ and by inductive assumption we therefore have $t'_kf(x'_k)+\sum\limits_{i=1}^{k-1} t_if(x_i)$ $ \leq f( t'_kx'_k+\sum\limits_{i=1}^{k-1} t_ix_i)=f(\sum\limits_{i=1}^{k+1} t_ix_i)$. To finish up, we need to show that $t_kf(x_k)+t_{k+1}f(x_{k+1}) \leq t'_kf(x'_k)$. Now, $0<t_k+t_{k+1}<1$ so there is some $r \in \mathbb{R}$ so that $rt_k+rt_{k+1}=1$. Then by the base case, $rt_kf(x_k)+rt_{k+1}f(x_{k+1}) \leq f(rt_kx_k+rt_{k+1}x_{k+1})=f(rt'_kx'_k)=f(x'_k)$. But by definition, $r= \frac{1}{t_k+t_{k+1}}$ and so dividing by $r$, this gives what we want.  
