If $a.b (a+b )=2000$ What is the value of $\frac {1}{a}+\frac {1}{b}+\frac {1}{a+b}$

Question

If $a\times b(a+b )=2000$ What is the value of $$\frac{1}{a}+\frac{1}{b}+\frac{1}{a+b}$$

By trial and error, I get the $a=10$,$b=10$ and so the answer should be $1/4$ , but is there any algebraic way to solve it?

Answer 1

$\frac{1}{a} + \frac{1}{b} + \frac{1}{a+b} = \frac{b(a+b) + a(a+b) + ab}{ab(a+b)}$.

I shall elaborate further. We can change the fraction into $\frac{(a+b)^2}{2000}+\frac{1}{a+b}$.

Note that since $ab(a+b)=2000$, $(a+b)=\frac{2000}{ab}$. We now substitute this into the above equation, getting $\frac{2000}{a^2b^2}+\frac{ab}{2000}$. This does not seem factorisable to me. Here, $a=10$ would definitely yield $\frac{1}{4}$, but the question is poorly defined with no single solution.

Answer 2

If we solve $ab^2+a^2b-2000=0$ for $b$ we obtain \begin{equation} b=\frac{-a^2\pm\sqrt{a(a^3+20^3)}}{2a} \end{equation} So \begin{equation} a+b=\frac{a^2\pm\sqrt{a(a^3+20^3)}}{2a} \end{equation} \begin{equation} \frac{1}{a+b}=\frac{2a}{a^2\pm\sqrt{a(a^3+20^3)}} \end{equation} which, when rationalized gives \begin{equation} \frac{1}{a+b}=\frac{2(-a^2\pm\sqrt{a(a^3+20^3)}}{20^3} \end{equation} By similiar steps you will find that \begin{equation} \frac{1}{b}=\frac{2(a^2\pm\sqrt{a(a^3+20^3)}}{20^3} \end{equation} So \begin{equation} \frac{1}{a}+\frac{1}{b}+\frac{1}{a+b}=\frac{1}{a}\pm \frac{4\sqrt{a(a^3+20^3)}}{20^3} \end{equation} If we let $a=10$ and use the positive branch the sum resolves to $\frac{1}{4}$, but for other values of $a$ one obtains different results for the sum.