Proof: Let the three sides of $\triangle ABC$ be $a, b, c$, and the area be $S$. Notice that
$$\begin{array}{l} \frac{r}{l_{A}}=\sin \frac{A}{2}=\sqrt{\frac{(p-b)(p-c)}{b c}} \\ \leqslant \frac{(p-b)+(p-c)}{2 \sqrt{b c}}=\frac{a}{2 \sqrt{b c}} \\ =\frac{a \sqrt{b c}}{2 b c} \leqslant \frac{a^{2}(b+c)}{4 a b c} . \\ \text { Then } \frac{3}{4}+\frac{r}{l_{A}}+\frac{r}{l_{B}}+\frac{r}{l_{C}} \\ \leqslant \frac{3}{4}+\frac{a^{2}(b+c)}{4 a b c}+\frac{b^{2}(c+a)}{4 a b c}+\frac{c^{2}(a+b)}{4 a b c} \\ =\frac{(a+b+c)(a b+b c+c a)}{4 a b c} . \\ \text { Also, }(a+b+c)^{2} \geqslant 3(a b+b c+c a), \\ \frac{3}{4}+\frac{r}{l_{A}}+\frac{r}{l_{B}}+\frac{r}{l_{C}} \leqslant \frac{(a+b+c)^{3}}{12 a b c} . \end{array}$$

Therefore, it suffices to prove $\frac{(a+b+c)^{3}}{12 a b c} \leqslant \frac{p^{2}}{12 r^{2}}$, which is equivalent to proving $8 p r^{2} \leqslant a b c$.
Notice that
$$8 p r^{2} \leqslant a b c \Leftrightarrow 8 r S \leqslant 4 R S \Leftrightarrow 2 r \leqslant R,$$

which holds by Euler's inequality. Thus, the original inequality is proven.