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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.

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    $\begingroup$ Just substitute b and c, you will get the required - "an equation in x y z and a". :-) $\endgroup$ – Mick Oct 6 '14 at 4:03
  • $\begingroup$ i think u misunderstood me.!! the question requires an answer like.. $y^2+z^2=15/4$ $\endgroup$ – maths lover Oct 6 '14 at 4:21
  • $\begingroup$ Yes, I know. It is just a joke. $\endgroup$ – Mick Oct 6 '14 at 4:29
  • $\begingroup$ anyone.????????? $\endgroup$ – maths lover Oct 6 '14 at 9:46
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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 $\frac{1}{a}=\frac{1}{\sqrt{b}}+\frac{1}{\sqrt{c}}$.

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  • $\begingroup$ it must be $15a^2/4$.!! sorry for the typo there.!! $\endgroup$ – maths lover Oct 18 '14 at 12:33
  • $\begingroup$ I don't see how that fixes things: $u$ should still tbe $\frac{1}{4}(-a^2+64a^4)$. $\endgroup$ – Kim Jong Un Oct 18 '14 at 15:44
  • $\begingroup$ try solving this eqn by just substituting x=0 $\endgroup$ – maths lover Oct 19 '14 at 6:12
  • $\begingroup$ I did solve it when $x=0$: did you not read the post above? $\endgroup$ – Kim Jong Un Oct 19 '14 at 8:28
  • $\begingroup$ can u tell an approach for reaching this answer?? $\endgroup$ – maths lover Oct 21 '14 at 15:22
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Partial answer: You need a simplification, not a solution.You already have an implicit equation in x, y, z and a ! The surface is symmetric about x = 0 or y-z plane ( change of sign before x makes no change). One needs to study (x,y) intersection only, as it is formed by rotation about x-axis ( due to appearance of $ y^2+z^2$ ).

EDIT: From the graphic, plane x=0 does not cut the surface along any real line of intersection.

It appears like, may be two (higher order?) spheres, centered at $ (\pm a/2,0,0)$ as seen in following Mathematica 3D plot. $a$ is taken as unity.

Plot_of_given_Equn

EDIT1:

Patience with algebraic simplification ( which is the solution of problem of seeing what you are dealing with!) is all that is needed.

$\sqrt{bc}/ ( \sqrt b + \sqrt c) = a $

$ (b c /a^2)^2 + (b+c)^2 - 2 b c (b+c)/ a^2 = b^2 c^2 $

Let $ ( b + c) / ( b c) = u $

$ u^2 + -2 u /a^2 + (1/a^4-1) $

$ u = 1/b + 1/c = 1/a^2 \pm 1 = 1/d $

So due to d there are two surfaces

$ 2 (x^2 + a^2/4 + y^2 + z^2) +2 b c = 1/d^2( (x^2 + a^2/4 + y^2 + z^2)^2 -( 2 a x)^2 $ which shows the two ordinary displaced fourth order spheres.

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