<s>[INST] How to prove $\frac{1}{x}=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+2\sqrt{\frac{1}{ab}+\frac{1}{ac}+\frac{1}{bc}}$ Question:

Let $a,b,c>0$ are give numbers and $x>0$, such that
  $$
\sqrt{\dfrac{a+b+c}{x}}=\sqrt{\dfrac{b+c+x}{a}}+\sqrt{\dfrac{c+a+x}{b}}+\sqrt{\dfrac{a+b+x}{c}}
$$
show that
  $$
\dfrac{1}{x}=\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}+2\sqrt{\dfrac{1}{ab}+\dfrac{1}{ac}+\dfrac{1}{bc}}
$$

I found this problem is very nice and here is my attempt:
Since
$$
\dfrac{a+b+c}{x}=\dfrac{b+c+x}{a}+\dfrac{c+a+x}{b}+\dfrac{a+b+x}{c}+2\sum_{cyc}\sqrt{\dfrac{(b+c+x)(c+a+x)}{ab}}
$$
it follows that
$$
\dfrac{a+b+c}{x}+3=\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)(a+b+c+x)+2\sum_{cyc}\sqrt{\dfrac{(b+c+x)(c+a+x)}{ab}}.
$$
I feel it's very ugly and I can't continue it.
Thank you.
 [/INST] You can prove this equation using geometry approach. Note that this equation is exactly similar with Descartes 4 Circles Theorem. Descartes' theorem says: If four circles are tangent to each other at six distinct points and the circles have curvatures $k_i$ (for $i = 1,\cdots, 4$), then $k_i$ satisfies the following relation:
$$
(k_1+k_2+k_3+k_4)^2=2(k_1^2+k_2^2+k_3^2+k_4^2),
$$
where  $k_i=\pm\dfrac{1}{r_i}$, $r_i$ is the radius of circle. The equation can also be written as:
$$
k_4=k_1+k_2+k_3\pm2\sqrt{k_1k_2+k_2k_3+k_1k_3},
$$
or
$$
\frac{1}{r_4}=\frac{1}{r_1}+\frac{1}{r_2}+\frac{1}{r_3}\pm2\sqrt{\frac{1}{r_1r_2}+\frac{1}{r_2r_3}+\frac{1}{r_1r_3}}.
$$
The generalization to $n$ dimensions or variables is referred to as the Soddy–Gosset theorem.
$$
\left(\sum_{i=1}^{n+2}k_i\right)^2=n\sum_{i=1}^{n+2}k_i^2.
$$
For detail explanation and complete proof of Descartes' theorem (also to answer your question), you may refer to these sites: 1, 2, or download this journal.
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$$\large\color{blue}{\text{# }\mathbb{Q.E.D.}\text{ #}}$$
</s>