HMMT November, 2023

Solution:
For any positive reals $a, b, c$, numbers $a, b, c$ are the side lengths of a triangle if and only if
$$(a+b+c)(-a+b+c)(a-b+c)(a+b-c)>0 \Longleftrightarrow \sum_{\mathrm{cyc}}\left(2 a^{2} b^{2}-a^{4}\right)>0$$
(to see why, just note that if $a \geq b+c$, then only the factor $-a+b+c$ is negative). Therefore, $x$ works if and only if
$$\begin{aligned} 2(x+7)(10-x)+2 x(x+7)+2 x(10-x) & >x^{2}+(x+7)^{2}+(10-x)^{2} \\ -5 x^{2}+46 x-9 & >0 \\ x & \in\left(\frac{1}{5}, 9\right) \end{aligned}$$
giving the answer $\frac{22}{25}$.

Solution:
Note that $\sqrt{x}<\sqrt{x+7}$, so $\sqrt{x}$ cannot be the maximum. Thus, $x$ works if and only if the following equivalent inequalities hold.
$$\begin{aligned} \sqrt{x} & >|\sqrt{x+7}-\sqrt{10-x}| \\ x & >(x+7)+(10-x)-2 \sqrt{(x+7)(10-x)} \\ \sqrt{4(x+7)(10-x)} & >17-x \\ 4(x+7)(10-x) & >x^{2}-34 x+289 \\ 4\left(70+3 x-x^{2}\right) & >x^{2}-34 x+289 \\ 0 & >5 x^{2}-46 x+9 \\ 0 & >(5 x-1)(x-9), \end{aligned}$$
so the range is $x \in\left(\frac{1}{5}, 9\right)$, and the answer is
$$\frac{9-\frac{1}{5}}{10}=\frac{22}{25}$$