<s>[INST] Solving $ \frac{1}{ a} = \ \frac{1}{ \ \sqrt{b}} \ +\ \frac{1}{ \ \sqrt{c}} $ with additional conditions How to solve this equation 
$$ \frac{1}{a}= \  \frac{1}{ \ \sqrt{b}} \ +\  \frac{1}{ \ \sqrt{c}} $$
where $$ b =  \sqrt{ (x-a/2)^2 + y^2 + z^2 )}$$
&   $$ c =  \sqrt{ (x+a/2)^2 + y^2 + z^2 )}$$
We have to get an equation in x y z and a!
I have tried to rationalize and do squaring but it becomes cumbersome.
Answer given is  $y^2+z^2=15/4$
I get this by putting x=0.
 [/INST] With the given info, the solution $y^2+z^2=\frac{15}{4}$ (or $\frac{15}{4}a^2$) is incorrect when $x=0$. One can verify the last statement by Mathematica. Here is another way to see it. 
By the AM-HM inequality, we have
$$
\frac{1}{\sqrt{b}}+\frac{1}{\sqrt{c}}\geq\frac{4}{\sqrt{b}+\sqrt{c}}.
$$
Now, use $x+y\leq\sqrt{2(x^2+y^2)}$ twice:
$$
\sqrt{b}+\sqrt{c}\leq\sqrt{2(b+c)}\leq\sqrt{2}(2b^2+2c^2)^{1/4}\leq\sqrt{2}[(a^2+4u+4x^2)]^{1/4}.
$$
Here $u=y^2+z^2\geq 0$. Putting this together, we get
$$
\frac{1}{a}\geq\frac{4}{\sqrt{2}[a^2+4u+4x^2]^{1/4}}\implies a^2+4u+4x^2\geq 64a^4.
$$
When $x=0$, all the inequalities above reduce to equalities, giving us
$$
a^2+4u=64a^4\implies u=\frac{1}{4}(-a^2+64a^4).
$$
Thus, when $x=0$, $u$ as given above for any $a$ (sufficiently large to ensure $u\geq 0$) will solve $\frac{1}{a}=\frac{1}{\sqrt{b}}+\frac{1}{\sqrt{c}}$.
</s>