$$\begin{array}{l} \left(\sqrt{a+\frac{1}{4}(b-c)^{2}}+\sqrt{b}+\sqrt{c}\right)^{2} \\ =\left(\sqrt{a+\frac{1}{4}(b-c)^{2}}+\frac{\sqrt{b}+\sqrt{c}}{2}+\frac{\sqrt{b}+\sqrt{c}}{2}\right)^{2} \\ \leq 3\left[a+\frac{1}{4}(b-c)^{2}+\left(\frac{\sqrt{b}+\sqrt{c}}{2}\right)^{2}+\left(\frac{\sqrt{b}+\sqrt{c}}{2}\right)^{2}\right] \\ =3\left[1-(b+c)+\frac{1}{4}(b-c)^{2}+\frac{1}{2}(b+c)+\sqrt{b c}\right] \\ =3\left[1-\frac{1}{2}(b+c)+\frac{1}{4}(b-c)^{2}+\sqrt{b c}\right] \\ =3\left[1-\frac{1}{2}(\sqrt{b}-\sqrt{c})^{2}+\frac{1}{4}(b-c)^{2}\right] \end{array}$$

To prove the inequality $(*)$, it is sufficient to prove:
$$1-\frac{1}{2}(\sqrt{b}-\sqrt{c})^{2}+\frac{1}{4}(b-c)^{2} \leq 1$$

which is equivalent to $-\frac{1}{2}(\sqrt{b}-\sqrt{c})^{2}+\frac{1}{4}(b-c)^{2} \leq 0$,
which in turn is equivalent to $(b-c)^{2}-2(\sqrt{b}-\sqrt{c})^{2} \leq 0$,
or $\quad(\sqrt{b}-\sqrt{c})^{2}\left[(\sqrt{b}+\sqrt{c})^{2}-2\right] \leq 0$.
Since $(\sqrt{b}+\sqrt{c})^{2}=b+c+2 \sqrt{b c} \leq 2(b+c)<2(a+b+c)=2$, the inequality $(**)$ holds, and thus the inequality $(*)$ is proved.