Show $f\left(\sum\limits_{k=1}^n \lambda_k x_k\right) \leq \sum\limits_{k=1}^n \lambda_k f(x_k)$ Let $n\in \mathbb{N}$ and $f:\mathbb{R}\rightarrow\mathbb{R}$
$$f(\sum\limits_{k=1}^n \lambda_k x_k) \leq \sum\limits_{k=1}^n \lambda_k f(x_k)$$
where $\sum\limits_{k=1}^n \lambda_k = 1$ and arbitrary $x_k \in \mathbb{R}$ and fixed $\lambda_k \in \mathbb{R} $ with the fixed sum of 1.
Do I understand that correctly, that $\sum\limits_{k=1}^n \lambda_k = 1$ for any $n$ or does it hold only for a specific $n$ and not necessarily $\sum\limits_{k=1}^{n+1} \lambda_k = 1$ for example?
Furthermore, I'd use induction to prove it, the start is trivial so I'll only show the step from n to n+1.
$$f(\sum\limits_{k=1}^{n+1} \lambda_k x_k)=f(\sum\limits_{k=1}^n \lambda_k x_k+\lambda_{n+1}x_{n+1})$$ 
but here's the problem because in order to use the hypothesis, I'd have to look at $f(\sum\limits_{k=1}^n \lambda_k x_k)$ alone, and I don't know how to "split it". I know I got to use the fact, that the function is convex, so it's increasing, continous and I got the defintion of a convex function. Can somebody give me a hint on what tool to use?
 A: You need to use both convexity and the induction hypothesis.
$$\begin{align}f(\sum\limits_{k=1}^{n+1} \lambda_k x_k)&=f(\sum\limits_{k=1}^n \lambda_k x_k+\lambda_{n+1}x_{n+1})\\
&=f\left[\left(\sum_{j=1}^n\lambda_j\right) \sum_{k=1}^n \frac{\lambda_k}{\sum_{j=1}^n\lambda_j}x_k + \lambda_{n+1}x_{n+1}\right]\\
&= f\left((1-\lambda_{n+1})\sum_{k=1}^n \frac{\lambda_k}{\sum_{j=1}^n\lambda_j}x_k + \lambda_{n+1}x_{n+1} \right) \\
&\leq (1-\lambda_{n+1})f\left(\sum_{k=1}^n \frac{\lambda_k}{\sum_{j=1}^n\lambda_j}x_k \right) + \lambda_{n+1}f(x_{n+1}) \\
&\leq \sum_{j=1}^n\lambda_j \sum_{k=1}^n \frac{\lambda_k}{\sum_{j=1}^n\lambda_j}f(x_k) + \lambda_{n+1}f(x_{n+1}) \\
&= \sum_{k=1}^{n+1}\lambda_k f(x_k)
\end{align}$$
When $\sum_{j=1}^n\lambda_j=0$, $\lambda_1=\ldots=\lambda_n=0$ and $\lambda_{n+1}=1$, so the clain holds.
A: When you say "fixed $\lambda_k$" it reminds me of this:  A function $f$ is called midpoint convex if
$$
f\left(\frac{x+y}{2}\right) \le \frac{f(x)+f(y)}{2}
$$
for all $x,y$.  In general (lacking the assumption of continuity), a midpoint convex function need not be convex.
