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.

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

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 $\frac{1}{a}=\frac{1}{\sqrt{b}}+\frac{1}{\sqrt{c}}$.

• it must be $15a^2/4$.!! sorry for the typo there.!! – maths lover Oct 18 '14 at 12:33
• I don't see how that fixes things: $u$ should still tbe $\frac{1}{4}(-a^2+64a^4)$. – Kim Jong Un Oct 18 '14 at 15:44
• try solving this eqn by just substituting x=0 – maths lover Oct 19 '14 at 6:12
• I did solve it when $x=0$: did you not read the post above? – Kim Jong Un Oct 19 '14 at 8:28
• can u tell an approach for reaching this answer?? – maths lover Oct 21 '14 at 15:22

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. 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.