Let $S$ be a nonempty convex set in $R^n$, and let $f: S \rightarrow R$. Show that $f$ is convex if and only if for any integer $k \geq 2$, the following holds true:
$x_1,...,x_k \in S$ implies that
$f(\sum\limits_{j=1}^k \lambda_j x_j) \leq \sum\limits_{j=1}^k \lambda_jf(x_j)$ , where $\sum\limits_{j=1}^k \lambda_j = 1, \lambda_j \geq 0$ for $j=1,..,k.$
So far I have managed to prove the inequality in forward implication when $f$ is assumed to be convex:
Suppose $f$ is convex on the nonempty convex set $S$ with $f: S \rightarrow R$. We proceed proving the implication by induction starting with the base case of $k = 2$. For $k = 2$ the relation is trivial and true by definition of convexity since for $x_1,x_2 \in S$ and for each $\lambda \in (0,1)$ $f$ being convex on $S$ implies
$f(\lambda x_1 + (1 - \lambda)x_2) \leq \lambda f(x_1) + (1 - \lambda)f(x_2)$
So suppose the relation holds for $n$, then we have
$f(\sum\limits_{j=1}^{n+1} \lambda_jx_j) = f(\sum\limits_{j=1}^n \lambda_jx_j + \lambda_{n+1}x_{n+1}) =$
$f(\lambda_{n+1}x_{n+1} + (1 - \lambda_{n+1})\frac{1}{1 - \lambda_{n+1}}\sum\limits_{j=1}^n \lambda_jx_j) \leq \lambda_{n+1}f(x_{n+1}) + (1 - \lambda_{n+1})f(\frac{1}{1 - \lambda_{n+1}} \sum\limits_{j=1}^n \lambda_jx_j) = $
$\lambda_{j+1}f(x_{n+1}) + (1 - \lambda_{n+1})f(\sum\limits_{j=1}^n \frac{\lambda_j}{1 - \lambda_{j+1}}x_j) \leq \lambda_{n+1}f(x_{n+1}) + (1 - \lambda_{n+1}) \sum\limits_{j=1}^n \frac{\lambda_j}{1 - \lambda_{j+1}}f(x_j) =$
$\sum\limits_{j=1}^n \lambda_j f(x_j) + \lambda_{n+1}x_{n+1} = \sum\limits_{j=1}^{n+1} \lambda_j f(x_j)$
So the relation holds for $n+1$ and by the principle of Mathematical Induction this holds for all $n \in N$
To complete the proof we now assume $x_1,...,x_k \in S$ implies that $f(\sum\limits_{j=1}^k \lambda_j x_j) \leq \sum\limits_{j=1}^k \lambda_jf(x_j)$ , where $\sum\limits_{j=1}^k \lambda_j = 1, \lambda_j \geq 0$ for $j=1,..,k.$
I am having trouble proceeding but have an idea of how to complete the proof for a given $n$. I could just construct an element that is the $n - 1$ sum of $\lambda_j x_j$ from 1 to $n$ i.e. show that $f(\sum\limits_{j=1}^{n} \lambda_jx_j) = f(\sum\limits_{j=1}^{n-1} \lambda_jx_j + \lambda_{n}x_{n}) \leq \sum\limits_{j=1}^{n-1} \lambda_j f(x_j) + \lambda_{n}x_{n} = \sum\limits_{j=1}^{n} \lambda_j f(x_j)$. Once this has been shown I could just modify it to fit the definition of convexity.
Or is there a more simple way to prove the only if backwards direction of the proof? Any help with this would be much appreciated, thanks.