Three circles of radii $$a$$, $$b$$, $$c$$ ($$a < b < c$$) touch each other externally. If they have $$x$$-axis as a common tangent, then:

First we note that each of the three circles is tangent to the $$x$$-axis from above. Hence the centre of the circle whose radius is $$a$$ can be written as $$O_{1}(x_{1},\,a)$$, the centre of the circle whose radius is $$b$$ as $$O_{2}(x_{2},\,b)$$ and that of the circle whose radius is $$c$$ as $$O_{3}(x_{3},\,c)$$.

Because all three circles touch one another externally, the distance between any two centres equals the sum of the corresponding radii. Using the distance formula $$\text{(distance)}^{2}=(\text{difference of abscissae})^{2}+(\text{difference of ordinates})^{2},$$ we can write three separate equations.

For the pair $$O_{1}O_{2}$$ we have

$$\bigl(x_{1}-x_{2}\bigr)^{2}+\bigl(a-b\bigr)^{2}=\bigl(a+b\bigr)^{2}.$$

For $$O_{1}O_{3}$$ we have

$$\bigl(x_{1}-x_{3}\bigr)^{2}+\bigl(a-c\bigr)^{2}=\bigl(a+c\bigr)^{2}.$$

For $$O_{2}O_{3}$$ we have

$$\bigl(x_{2}-x_{3}\bigr)^{2}+\bigl(b-c\bigr)^{2}=\bigl(b+c\bigr)^{2}.$$

From the first equation we obtain

$$\bigl(x_{1}-x_{2}\bigr)^{2}=(a+b)^{2}-(b-a)^{2}=4ab,$$ so $$|x_{1}-x_{2}|=2\sqrt{ab}.$$

The second equation gives

$$\bigl(x_{1}-x_{3}\bigr)^{2}=(a+c)^{2}-(c-a)^{2}=4ac,$$ so $$|x_{1}-x_{3}|=2\sqrt{ac}.$$

The third gives

$$\bigl(x_{2}-x_{3}\bigr)^{2}=(b+c)^{2}-(c-b)^{2}=4bc,$$ so $$|x_{2}-x_{3}|=2\sqrt{bc}.$$

Because $$a<b<c,$$ the smallest circle is most naturally sandwiched between the other two on the common tangent. We therefore assume the left-to-right order

$$x_{2}\;<\;x_{1}\;<\;x_{3},$$

so that

$$x_{1}-x_{2}=2\sqrt{ab},\qquad x_{3}-x_{1}=2\sqrt{ac},\qquad x_{3}-x_{2}=2\sqrt{bc}.$$

Now we use the simple relation of collinear points on a line:

$$x_{3}-x_{2}=(x_{3}-x_{1})+(x_{1}-x_{2}).$$

Substituting the expressions just found, we get

$$2\sqrt{bc}=2\sqrt{ac}+2\sqrt{ab}.$$

Dividing every term by $$2\sqrt{abc}$$ simplifies the equation:

$$\frac{\sqrt{bc}}{\sqrt{abc}}=\frac{\sqrt{ac}}{\sqrt{abc}}+\frac{\sqrt{ab}}{\sqrt{abc}}.$$

Simplifying each fraction separately, we find

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

This is exactly the relation stated in Option A, while none of the other options follow from the geometry. Hence, the correct answer is Option A.