HMMT November

Solution:
Answer: $\frac{23}{8}=2.875$

Let $a=\sqrt{3-x}$, $b=\sqrt{4-x}$, $c=\sqrt{6-x}$, so $x=ab+bc+ca$.

Then
$$(a+b)(a+c)=a^2+ab+bc+ca=(3-x)+x=3$$
Likewise,
$$(b+a)(b+c)=4$$
$$(c+a)(c+b)=6$$
By multiplying these equations and taking the square root, we get that
$$(a+b)(b+c)(c+a)=\sqrt{72}=6\sqrt{2}$$
so
$$b+c=2\sqrt{2}, \quad c+a=\frac{3\sqrt{2}}{2}, \quad a+b=\sqrt{2}$$
and hence
$$2a=\frac{3\sqrt{2}}{2}+\sqrt{2}-2\sqrt{2} \Longrightarrow a=\frac{\sqrt{2}}{4}.$$
Since $a=\sqrt{3-x}$, it follows that $x=\frac{23}{8}$.

Solution:
The given equation implies
$$(\sqrt{3-x}+\sqrt{4-x}+\sqrt{6-x})^{2} = (3-x)+(4-x)+(6-x) + 2(\sqrt{(3-x)(4-x)}+\sqrt{(4-x)(6-x)}+\sqrt{(6-x)(3-x)}) = 13-3x+2x=13-x$$
As $\sqrt{3-x}+\sqrt{4-x}+\sqrt{6-x}$ is nonnegative,
$$\sqrt{3-x}+\sqrt{4-x}+\sqrt{6-x}=\sqrt{13-x}$$
We repeatedly rearrange, square both sides, and simplify:
$$\begin{aligned} \sqrt{3-x}+\sqrt{4-x} &= \sqrt{13-x}-\sqrt{6-x} \\ 7-2x+2\sqrt{(3-x)(4-x)} &= 19-2x-2\sqrt{(13-x)(6-x)} \\ \sqrt{(3-x)(4-x)} &= 6-\sqrt{(13-x)(6-x)} \\ (3-x)(4-x) &= 36-12\sqrt{(13-x)(6-x)}+(13-x)(6-x) \\ 2x-17 &= 2\sqrt{(13-x)(6-x)} \\ 4x^{2}-68x+289 &= 4(13-x)(6-x) \\ 8x &= 23 \Longrightarrow x=\frac{23}{8}. \end{aligned}$$