The proof methods for this problem are numerous, but the author only uses the idea elaborated above: from $(a+2)^{2}=(b+2)^{2}=\frac{25}{4}$, we conjecture that the condition for equality is $a=b=\frac{1}{2}$. Verifying this conjecture: using the inequality $(a+2)^{2}+\frac{25}{4} \geqslant 5(a+2)$, similarly $(b+2)^{2}+\frac{25}{4} \geqslant 5(b+2)$, the equality holds precisely when $a=b=\frac{1}{2}$,
$$\therefore(a+2)^{2}+(b+2)^{2}+\frac{25}{2} \geqslant 5(a+b+4)=$$
25, thus we can prove that $(a+2)^{2}+(b+2)^{2} \geqslant \frac{25}{2}$.