Prove Jensen's Inequality: Convex Functions (a,b) → R

[v.rad]

1. Suppose that f: (a,b) --> R is convex. Prove Jensen's inequality: if x₁,...,x_n$$\in(a,b)$$ and c₁,...,c_n >= 0 s.t. $$\sum(c_j)f(x_j)$$ >= f($$\sum((c_j)(x_j))$$

both summations from j = 1 to n

2: Convex: whenever x₁, x₂ $$\in(a,b)$$ and 0 <= c <= 1, we have cf(x₁) + (1 + c)f(x₂) >= f(cx₁ + (1-c)x₂)


3. I realize that this is a proof by induction.
I have written the first summation in the form of (c_n+1)(x_n+1) + (1 - (c_n+1)y, and then used my definition of convexity. My issue is trying to figure out what my y is. I know it has something to do with breaking it into a summation from just j=1 to n but I keep going in circles...

 

[jbunniii]

Try this:

$$\sum_{j=1}^{n+1}c_j x_j = c_{n+1} x_{n+1} + \sum_{j=1}^{n}c_j x_j = c_{n+1} x_{n+1} + \left(\frac{1 - c_{n+1}}{1 - c_{n+1}}\right) \sum_{j=1}^{n}c_j x_j = c_{n+1} x_{n+1} + (1 - c_{n+1}) \sum_{j=1}^{n}\frac{c_j}{1 - c_{n+1}} x_j$$

What is

$$\sum_{j=1}^{n}\frac{c_j}{1 - c_{n+1}}$$?