How do I rewrite this function in terms of two other function?

Question

I'm trying to establish a relationship between these three terms

$$U(x) = \frac{a(x_1+x_2)}{x_1x_2}$$ $$g(x) = \frac{ax_2}{x_1(x_1+x_2)}$$ $$f(x) = \frac{ax_1}{x_2(x_1+x_2)}$$

How do I write $U(x)$ in terms of the other two functions?

Answer 1

Hint: Eliminate $$x_1,x_2$$ from the System $$x_1^2+x_1x_2-\frac{ax_2}{g}=0$$ $$x_1x_2+x_2^2-\frac{a}{f}x_1=0$$

Answer 2

$$ f g = \frac{a^2 x_1 x_2}{(x_1+x_2)^2 x_1 x_2} = \frac{a^2}{(x_1+x_2)^2}\\ g+f = \frac{ax_2^2+ax_1^2}{x_1 x_2 (x_1 + x_2)}\\ $$

Define the following functions and give the Newton identities for them

$$ e_1 = p_1 = x_1 + x_2\\ p_2 = x_1^2 + x_2^2\\ e_2 = x_1 x_2\\\ 2 e_2 = e_1 p_1 - p_2 $$

So we have

$$ fg = \frac{a^2}{p_1^2}\\ f+g=\frac{ap_2}{e_2 p_1} = \frac{a e_1 p_1 - 2 a e_2}{e_2 p_1}\\ = \frac{a e_1}{e_2} - \frac{2a}{p_1} = \frac{a e_1}{e_2} - 2 \sqrt{fg}\\ U = \frac{a e_1}{e_2} = (f+g)+2\sqrt{fg} $$

Proving the required Newton identity is

$$ 2 x_1 x_2 = (x_1 + x_2)^2 - (x_1^2+x_2^2)\\ $$

The reason one recognizes to use this strategy is the fact that the goal $U$ is a rational function of $x_1 , x_2$ and symmetric between them.