I'm having trouble with the following task and would appreciate some help with solving it, any help would be much appreciated.
Let: $$x_1=\frac{m}{p_1+\sqrt{p_1·p_2}}$$ $$x_2=\frac{m}{\sqrt{p_1·p_2}+p_2}$$
Show that: $$-(\frac{1}{x_1}+\frac{1}{x_2})=-(\sqrt{p_1}+\sqrt{p_2})^2·\frac{1}{m}$$
Thanks in advance for any help, Mathias
Your suggested solution is true if only $p_1$ AND $p_2$ are both positive. (As @dxiv mentioned) Otherwise, you should include abstract under the radicals.
i.e. $\displaystyle (\frac {1}{x_1} + \frac {1}{x_2})=\frac{-1}{m}(\sqrt {|p_1|}+\sqrt{|p_2|})^2$
Here is the solution:
$\displaystyle x_1=\frac{m}{p_1+\sqrt{p_1·p_2}} \to \frac{1}{x_1}=\frac{p_1+\sqrt{p_1·p_2}}{m}$
$\displaystyle x_2=\frac{m}{p_2+\sqrt{p_1·p_2}} \to \frac{1}{x_2}=\frac{p_2+\sqrt{p_1·p_2}}{m}$
$\displaystyle \to \frac {1}{x_1} + \frac {1}{x_2}=\frac{p_1+\sqrt{p_1·p_2}}{m} + \frac{p_2+\sqrt{p_1·p_2}}{m} = \frac{p_1+p_2+2\sqrt{p_1p_2}}{m}=\frac{\sqrt {p_1^2}+\sqrt{p_2^2}+2\sqrt{p_1p_2}}{m}=\frac{(\sqrt {p_1}+\sqrt{p_2})^2}{m}$
$\displaystyle \to -(\frac {1}{x_1} + \frac {1}{x_2})=\frac{-1}{m}(\sqrt {p_1}+\sqrt{p_2})^2$