Prove that$ f\Bigg(\sum_{i=1}^n \lambda_i x_i\Bigg) \leq \sum_{i=1}^n \lambda_i f(x_i) $ for a convex function "f"

Question

Let f be a convex function defined on a set I. If $x_1, x_2, ...,x_n \in I$, and $\lambda_1, \lambda_2 ,...,\lambda_n \in [0,1]$ with $\sum_{i=1}^n \lambda_i=1$ then prove that $$ f\Bigg(\sum_{i=1}^n \lambda_i x_i\Bigg) \leq \sum_{i=1}^n \lambda_i f(x_i) $$

I have no idea where to start, except to use the definition of a convex function: $$ f(\lambda x+(1-\lambda)y) \leq \lambda f(x) + (1-\lambda)f(y) $$

Anyone any ideas of how to approach this problem?

Answer 1

For $n = 2$ you know it's true. Let's proceed by induction on $n$: let $$\tau_i = \frac{\lambda_i}{\sum\limits_{j=1}^{n-1} \lambda_j}$$

Then:

\begin{align*} f\left(\sum_{j=1}^n \lambda_j x_j\right) &= f\left(\left(\sum_{j=1}^{n-1}\lambda_j\right)\cdot\sum_{j=1}^{n-1}\tau_jx_j + \lambda_nx_n\right) \\ &\leq\left(\sum_{j=1}^{n-1}\lambda_j\right)\cdot f\left(\sum_{j=1}^{n-1} \tau_jx_j\right) + \lambda_nf(x_n) \\ &\leq\left(\sum_{j=1}^{n-1}\lambda_j\right)\cdot \sum_{j=1}^{n-1} \tau_jf(x_j) + \lambda_n f(x_n) = \sum_{j=1}^n \lambda_j f(x_j) \end{align*}

Where the second inequality comes from the induction hypothesis and the first from the property for $n=2$.