If abc

=>
\(\sqrt{a+b}+\sqrt{b+c}=\sqrt{c+a}\) \(\left(\sqrt{a+b}+\sqrt{b+c}\right)^2=\left(\sqrt{c+a}\right)^2\) (squaring both sides)
\(\left(\sqrt{a+b}+\sqrt{b+c}\right)\left(\sqrt{a+b}+\sqrt{b+c}\right)=c+a\) (multiplying)
a + b + √((a + b)(b + c)) + b + c = c + a (foiling)
a + 2b + c + 2√((a + b)(b + c)) = c + a (adding like terms)
2√((a + b)(b + c)) = -2b (subtracting a + 2b + c from both sides)
√((a + b)(b + c)) = -b (dividing both sides by 2)
(a + b)(b + c) = b^2 (squaring both sides)
ab + ac + b^2 + bc = b^2 (foiling)
ab + ac + bc = 0 (subtracting b^2 from both sides)
ab/abc + ac/abc + bc/abc = 0/abc (dividing both sides by abc)
1/a + 1/b + 1/a = 0 (simplifying)

Therefore, A is the answer.
Answer: A