Let $f(x)=x^2+3x-3,x\gt0.$ If $n$ points $x_1,x_2,...,x_n$ are chosen on the x-axis, evaluate $\frac{x_1+x_2+...+x_n}n$

Question

Question:

Let $f(x)=x^2+3x-3,x\gt0.$ If $n$ points $x_1,x_2,...,x_n$ are so chosen on the x-axis such that

(i) $\frac1n\sum_{i=1}^nf^{-1}(x_i)=f(\frac1n\sum_{i=1}^nx_i)$

(ii) $\sum_{i=1}^nf^{-1}(x_i)=\sum_{i=1}^nx_i$

Evaluate $\frac{x_1+x_2+...+x_n}n$

My Attempt:

I found $f^{-1}(x)=\frac{-3+\sqrt{21+x}}2$

But I don't think it is required.

We need to find $\frac{\sum_{i=1}^nx_i}n$, which is same as $\frac{\sum_{i=1}^nf^{-1}(x_i)}n$, which is same as $f(\frac1n\sum_{i=1}^nx_i)$

I am thinking of Jensen's inequality. But don't know how to apply that here.

Answer 1

Combining $(i)$ and $(ii)$, one has directly $$\dfrac{1}{n} \sum_{i=1}^n x_i = f\left(\frac1n\sum_{i=1}^nx_i\right)$$

So $\displaystyle \dfrac{1}{n} \sum_{i=1}^n x_i$ satisfies the equation $x=f(x)$, i.e. $x^2 + 3x - 3 = x$, i.e. $(x-1)(x+3)=0$.

Since $\displaystyle \dfrac{1}{n} \sum_{i=1}^n x_i >0$, then the only possibility is that $$\boxed{\dfrac{1}{n} \sum_{i=1}^n x_i=1}$$