Prove Monotonically Decreasing Convex Function Boundary

[stainburg]

Let \(\displaystyle f\):\(\displaystyle (0,\infty) \to (0,\infty)\) be a monotonically decreasing convex function. Choose \(\displaystyle 0<x_1<x_2<...<x_n\), and \(\displaystyle \lambda_i \geq 0\) such that \(\displaystyle \sum_{i=0}^{n}\lambda_i=1\). Prove that

\(\displaystyle \sum_{j=1}^{n}\lambda_j x_j\sum_{i=1}^{n}\lambda_i f(x_i) \leq \frac{(x_nf(x_1)-x_1f(x_n))^2}{4(x_n-x_1)(f(x_1)-f(x_n))}\) .

 

[jvicens]

Thank you for your post. I would like to provide a proof for the inequality you have stated.

First, note that since f is a monotonically decreasing convex function, it is also a continuous function on the interval (0, ∞). This means that it satisfies the Intermediate Value Theorem, which states that for any two points x_1 and x_2 in the domain of f, there exists a point c in between them such that f(c) is equal to the average of f(x_1) and f(x_2), i.e. f(c) = (f(x_1) + f(x_2))/2.

Now, let us consider the function g(x) = x[f(x_1) - f(x_n)] - f(x_1)x_n + f(x_n)x_1. This function is also continuous on the interval (0, ∞) and can be rewritten as g(x) = (x - x_n)(f(x_1) - f(x_n)) + (x_n - x_1)f(x_n). It is clear that g(x_1) = g(x_n) = 0.

Next, let us consider the function h(x) = g(x)/x. Since x > 0 for all x in the domain, h(x) is also continuous on (0, ∞) and we have h(x_1) = h(x_n) = 0. Furthermore, h(x) is also differentiable on (0, ∞) and we have h'(x) = (x[f'(x_1) - f'(x_n)] + f(x_1) - f(x_n))/x^2.

Now, let us define the function F(x) = f(x) - [f(x_1) - f(x_n)](x - x_n)/(x_n - x_1). Since f is convex, it follows that f'(x) \geq [f(x_1) - f(x_n)]/(x_n - x_1), which implies that F'(x) \geq 0 for all x \in (0, ∞). Therefore, F(x) is a monotonically increasing function on (0, ∞) and we have F(x_1) = f(x_1) and F(x_n) = f(x_n).

Now, let