Proof: Without loss of generality, let $a=\max \{a, b, c\}$. From the given condition, we have $a^{2} \leqslant b^{2}+c^{2}$. Then
(1)
$$\begin{array}{l} \Leftrightarrow \frac{2-2 b^{2}}{2 a}+\frac{2-2 c^{2}}{2 b}+\frac{2-2 a^{2}}{2 c} \leqslant \frac{5}{4} \\ \Leftrightarrow \frac{a^{2}+c^{2}-b^{2}}{2 a}+\frac{a^{2}+b^{2}-c^{2}}{2 b}+\frac{b^{2}+c^{2}-a^{2}}{2 c} \leqslant \frac{5}{4} . \end{array}$$

And
$$\frac{5}{4}=\frac{5}{4} \sqrt{\frac{1}{2}\left(a^{2}+b^{2}+c^{2}\right)} \geqslant \frac{5}{4} \cdot \frac{1}{2}\left(a+\sqrt{b^{2}+c^{2}}\right),$$

Therefore, it suffices to prove
$$\begin{aligned} & \frac{a^{2}+c^{2}-b^{2}}{a}+\frac{a^{2}+b^{2}-c^{2}}{b}+\frac{b^{2}+c^{2}-a^{2}}{c} \\ \leqslant & \frac{5}{4}\left(a+\sqrt{b^{2}+c^{2}}\right) . \\ \Leftrightarrow & \frac{b c\left(a^{2}+c^{2}-b^{2}\right)+c a\left(a^{2}+b^{2}-c^{2}\right)+a b\left(b^{2}+c^{2}-a^{2}\right)}{a b c} \\ \leqslant & \frac{5}{4}\left(a+\sqrt{b^{2}+c^{2}}\right) \\ \Leftrightarrow & a+b+c+\frac{(a+b+c)(a-b)(b-c)(c-a)}{a b c} \\ \leqslant & \frac{5}{4}\left(a+\sqrt{b^{2}+c^{2}}\right) \\ \Leftrightarrow & 4(a+b+c)[a b c+(a-b)(b-c)(c-a)] \\ \leqslant & 5 a b c\left(a+\sqrt{b^{2}+c^{2}}\right) \\ \Leftrightarrow & 4(a+b+c)(a-b)(a-c)(c-b)+a b c(4(b+c) \\ & \left.-a-5 \sqrt{b^{2}+c^{2}}\right) \leqslant 0 . \end{aligned}$$
$$\begin{array}{l} \text { Let } f(a)=4(a+b+c)(a-b)(a-c)(c-b)+a b c(4(b+c)- \\ \left.a-5 \sqrt{b^{2}+c^{2}}\right) . \end{array}$$
(i) If $b \leqslant c \leqslant a \leqslant \sqrt{b^{2}+c^{2}}$. When $b=c$, $a \leqslant \sqrt{2} b$. At this time, $f(a)=$ $a b^{2}(8 b-a-5 \sqrt{2} b)=a b^{2}((7-5 \sqrt{2}) b+(b-a)) \leqslant 0$, hence $\lim _{a \rightarrow-\infty} f(a) \leqslant 0$, and $f(0)=4(b+c) b c(c-b)>0$, $f(c)=b c^{2}\left(4(b+c)-c-5 \sqrt{b^{2}+c^{2}}\right)=b c^{2}(4 b+3 c - 5 \sqrt{b^{2}+c^{2}}) \leqslant 0 \Leftrightarrow (3 b-4 c)^{2}>0$. Therefore, $f(a)$ has three real roots in $(-\infty, 0)$, $(0, c)$, and $(c,+\infty)$.

Consider the sign of $f\left(\sqrt{b^{2}+c^{2}}\right)$, noting that the sign of $f(a)$ is the same as the sign of the left side minus the right side in (3). We only need to consider
$$\frac{a^{2}+c^{2}-b^{2}}{a}+\frac{a^{2}+b^{2}-c^{2}}{b}+\frac{b^{2}+c^{2}-a^{2}}{c}-\frac{5}{4}\left(a+\sqrt{b^{2}+c^{2}}\right)$$

when $a=\sqrt{b^{2}+c^{2}}$. At this time,
$$\text { (4) } \begin{aligned} & =\frac{2 c^{2}}{\sqrt{b^{2}+c^{2}}}+\frac{2 b^{2}}{b}-\frac{5}{4}\left(\sqrt{b^{2}+c^{2}}+\sqrt{b^{2}+c^{2}}\right) \\ & =\frac{2 c^{2}}{\sqrt{b^{2}+c^{2}}}+2 b-\frac{5}{2} \sqrt{b^{2}+c^{2}} \\ & =\frac{1}{2 \sqrt{b^{2}+c^{2}}}\left(4 c^{2}+4 b \sqrt{b^{2}+c^{2}}-5\left(b^{2}+c^{2}\right)\right) \\ & =\frac{1}{2 \sqrt{b^{2}+c^{2}}}\left(4 b \sqrt{b^{2}+c^{2}}-5 b^{2}-c^{2}\right) \leqslant 0 \end{aligned}$$
$\left(\Leftrightarrow 4 b \sqrt{b^{2}+c^{2}} \leqslant 5 b^{2}+c^{2} \Leftrightarrow 16 b^{4}+16 b^{2} c^{2} \leqslant 25 b^{4}+c^{4}+10 b^{2} c^{2} \Leftrightarrow 9 b^{4}+c^{4}-6 b^{2} c^{2} \geqslant 0 \Leftrightarrow (3 b^{2}-c^{2})^{2} \geqslant 0\right)$.

From the above, we know that $f\left(\sqrt{b^{2}+c^{2}}\right) \leqslant 0$. This indicates that when $c \leqslant a \leqslant \sqrt{b^{2}+c^{2}}$, $f(a) \leqslant 0$.
(ii) If $c < b \leqslant a \leqslant \sqrt{b^{2}+c^{2}}$, then $a > c$ and $a > b$, so
$$\begin{aligned} & 4(a+b+c)(a-c)(a-b)(b-c) \geqslant 0 \\ \geqslant & 4(a+b+c)(a-c)(a-b)(c-b), \end{aligned}$$

Therefore, when $b \leqslant a \leqslant \sqrt{b^{2}+c^{2}}$,
$$\begin{array}{c} f(a) \leqslant 4(a+b+c)(a-c)(a-b)(b-c)+ \\ a b c\left(4(b+c)-a-5 \sqrt{b^{2}+c^{2}}\right) \leqslant 0 . \end{array}$$
$\left(a=1, b=\frac{1}{2}, c=\frac{\sqrt{3}}{2}\right.$ when equality holds. $)$