<s>[INST] Find the value of $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$ 
Suppose $(a, b, c)\in\Bbb R^3$, $a,b,c$ are all nonzero, and we have $
\sqrt{a+b}+\sqrt{b+c}=\sqrt{c+a}$. Find the value of $$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}.$$

Here is my attempt: 
$$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{bc+ac+ab}{abc}.$$
I am having trouble in figuring out the best approach to simplify $
\sqrt{a+b}+\sqrt{b+c}=\sqrt{c+a}$
so that I can find the value of $
\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$ . Hope somebody has an idea.
 [/INST] Take the square of the condition :
$$a+b+b+c+2\sqrt{a+b}\sqrt{b+c}=a+c$$
This implies $b=-\sqrt{a+b}\sqrt{b+c}$, so $b^2=(a+b)(b+c)$, hence
$$ab+bc+ac=0$$
So your answer should be $0$ (what else could it be ?) :-)
EDIT : 
I thought about this a little (well, OK, a big !) more, and especially the feasibility of the initial condition.
We have $(1)\,:\,ab+ac+bc=0$, so $a$, $b$ and $c$ can't be all positive. Which asks the question : can $a+b$, $a+c$ and $b+c$ be positive ?
Now we can't have two negative values, because the sum would be negative. So only one of the three variables can be negative. And it can't be $a$ or $c$, because, for example, if $a<0$, $a+c<c<b+c$, and the left member of the condition is strictly greater than the right member.
So $b<0$, and solving for $b$ in $(1)$, we find $b=-\frac{ac}{a+c}=-\frac{1}{1/a+1/c}$ (which, BTW, is half the opposite of the harmonic mean of $a$ and $c$).
Remains to prove that $a+b$ and $b+c$ are both positive. For example
$$a-\frac{ac}{a+c}=\frac{a^2}{a+c}>0$$
Now I can sleep without thinking anymore at this funny problem :-)
END EDIT.
And BTW : welcome to MSE !
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