How can I prove this kind of reverse Jensen's inequality?

Question

$$f(\lambda_1x_1+...+\lambda_nx_n)\ge\lambda_1f(x_1)+...+\lambda_nf(x_n)$$

f is a convex function and Conditions are as follows: $$\lambda_1\gt0 , \lambda_i\le0, \lambda_1+...+\lambda_n=1$$

Answer 1

Given the conditions you actually have $\lambda_1 = 1+ \sum_{j\geq 2} (-\lambda_j)\geq 1$ or $1= \frac{1}{\lambda_1} + \sum_{j\geq 2} \frac{-\lambda_j}{\lambda_1}$ (sum of non-negative elements). Write $x_0=\sum_{j\geq 1}\lambda_j x_j$. Then $$ x_1 = \frac{1}{\lambda_1} x_0 + \sum_{j\geq 2} \frac{-\lambda_j}{\lambda_1} x_j $$ and by Jensen (usual): $$f(x_1) \leq \frac{1}{\lambda_1} f(x_0) + \sum_{j\geq 2} \frac{-\lambda_j}{\lambda_1} f(x_j)$$