5. In $\triangle A B C$, the three sides are $a, b, c$, and the lengths of the medians to sides $B C, C A, A B$ are $m_{a}, m_{b}, m_{c}$, respectively. Then the minimum value of $\frac{m_{a}^{2}}{a^{2}}+\frac{m_{b}^{2}}{b^{2}}+\frac{m_{c}^{2}}{c^{2}}$ is $\qquad$.
By the median length formula $4 m_{a}^{2}+a^{2}=2\left(b^{2}+c^{2}\right) \Rightarrow \frac{m_{a}^{2}}{a^{2}}=\frac{2\left(b^{2}+c^{2}\right)-a^{2}}{4 a^{2}}=\frac{1}{2}\left(\frac{b^{2}}{a^{2}}+\frac{c^{2}}{a^{2}}\right)-\frac{1}{4}$, then $\frac{m_{a}^{2}}{a^{2}}+\frac{m_{b}^{2}}{b^{2}}+\frac{m_{c}^{2}}{c^{2}}=\frac{1}{2}\left(\frac{b^{2}}{a^{2}}+\frac{c^{2}}{a^{2}}+\frac{c^{2}}{b^{2}}+\frac{a^{2}}{b^{2}}+\frac{a^{2}}{c^{2}}+\frac{b^{2}}{c^{2}}\right)-\frac{3}{4} \geqslant \frac{1}{2} \times 6-\frac{3}{4}=\frac{9}{4}$, equality holds when $a=b=c$. Therefore, the minimum value of $\frac{m_{a}^{2}}{a^{2}}+\frac{m_{b}^{2}}{b^{2}}+\frac{m_{c}^{2}}{c^{2}}$ is $\frac{9}{4}$.