Show a certain point of a function is a product of the function itself at different points

Question

Question: Suppose the function $f:(a,b)\rightarrow \mathbb{R}$ is continuous on $(a,b)$ and $f(x)>0$ for any $x \in (a,b)$. Given the $x_1<x_2<...<x_n \in (a,b)$ and the positive real numbers $\alpha_1 +\alpha_2 + ... + \alpha_n = 1$, show that there exists $x_0 \in (a,b)$ such that $f(x_0) = f(x_1)^{\alpha_1} \times f(x_2)^{\alpha_2} \times ... f(x_n)^{\alpha_n}$ holds.

My approach: I start with applying the Extreme Value Theorem, which implies that there exists $x_0 \in (a,b)$ such that $f(x_0) \leq f(x)$ for any $x \in (a,b)$. Clearly, $f(x_0) \leq f(x_i)$ for $i = 1,2,...,n$. Then multiply them all and we get $(f(x_0))^n \leq f(x_1) f(x_2)...f(x_n)$. Now taking the nth root and using $f(x) > 0$ for $x \in (a,b)$, $f(x_0) \leq f(x_1)^{1/n} f(x_2)^{1/n}...f(x_n)^{1/n}$. It is obvious that $n > 1$ or the question will be just a nonsense, so $\frac{1}{n} < 1$. Then clearly there exists positive real numbers $\alpha_1,\alpha_2 , ... , \alpha_n < 1$ such that $f(x_0) = f(x_1)^{\alpha_1} f(x_2)^{\alpha_2}...f(x_n)^{\alpha_n}$. Now my problem is proving $\alpha_1 +\alpha_2 + ... + \alpha_n = 1$. My approach might be wrong and I cannot think of a way to show the sum equals to 1. Any help will be greatly appreciated!

Answer 1

Note first the problem is trivial if $f$ is a constant. Now since $f:[x_1,x_n]\to\mathbb{R}^+$ is continuous then it attains a maximum and minimum at different values. WLOG let $x_1\le\alpha<\beta\le x_n$, such that $f(\alpha)\le f(x) \le f(\beta)$ for all $x\in[x_1,x_n]$.

We see then that

$$f(\alpha) = f(\alpha)^{\alpha_1}\cdots f(\alpha)^{\alpha_n} \le f(x_1)^{\alpha_1}\cdots f(x_n)^{\alpha_n} \le f(\beta)^{\alpha_1}\cdots f(\beta)^{\alpha_n} = f(\beta)$$

since $\alpha_1+\cdots+\alpha_n=1$. By the intermediate value theorem, there must exist a number in $x_0\in[\alpha,\beta]$ such that

$$f(x_0)=f(x_1)^{\alpha_1}\cdots f(x_n)^{\alpha_n}$$