IMO 2001: a/√(a² + 8bc)

Weighted Jensen. For a convex $f$ and weights $w_i\ge0$ with $\sum w_i=1$: $\sum w_if(t_i)\ge f\left(\sum w_it_i\right)$. Take $f(t)=t^{-1/2}$, which is convex on $(0,\infty)$ since $f''(t)=\frac34t^{-5/2}>0$, with $w_a=\frac as$ and $t_a=X_a$:
$$\sum\frac as\cdot\frac1{\sqrt{X_a}}\ge\frac1{\sqrt{\sum\frac as X_a}}.$$
Multiplying by $s$,
$$\sum\frac a{\sqrt{X_a}}\ge\frac{s^{3/2}}{\sqrt{\sum aX_a}}.$$

It suffices to prove $s^3\ge\sum aX_a=a^3+b^3+c^3+24abc$. Now
$$s^3-\left(\sum a^3+24abc\right)=3\left(\sum_{\rm sym}a^2b-6abc\right)\ge0,$$
because $\sum_{\rm sym}a^2b\ge6abc$.

Equality iff $a=b=c$.

Remark. The official solutions use Hölder's inequality or the bound $\frac a{\sqrt{a^2+8bc}}\ge\frac{a^{4/3}}{a^{4/3}+b^{4/3}+c^{4/3}}$.

Lesson. A sum $\sum\frac a{\sqrt{X_a}}$ with the *numerator also a weight* suggests Jensen with weights proportional to the numerators.