Proof: According to the Cauchy-Schwarz inequality, we have
$$\begin{array}{l} \left(\sqrt{3 a^{2}+1}+\sqrt{5 a^{2}+3 b^{2}+1}+\sqrt{7 a^{2}+5 b^{2}+3 c^{2}+1}\right)^{2} \\ =\left(\frac{1}{\sqrt{6}} \sqrt{6\left(3 a^{2}+1\right)}+\frac{1}{\sqrt{4}} \sqrt{4\left(5 a^{2}+3 b^{2}+1\right)}+\frac{1}{\sqrt{3}} \sqrt{3\left(7 a^{2}+5 b^{2}+3 c^{2}+1\right)}\right)^{2} \\ \leq\left(\frac{1}{6}+\frac{1}{4}+\frac{1}{2}\right)\left(6\left(3 a^{2}+1\right)+4\left(5 a^{2}+3 b^{2}+1\right)+3\left(7 a^{2}+5 b^{2}+3 c^{2}+1\right)\right)^{2} \end{array}$$

Thus, it suffices to prove
$$59 a^{2}+27 b^{2}+9 c^{2} \geq 95$$

Noting that $a \leq b \leq c$, we have
$$a b+b c+c a \geq 2 a b+b^{2} \geq 2 a^{2}+b^{2}$$

or
$$5 a^{2}+3 b^{2}+c^{2} \leq(a+b+c)^{2}=9 \Rightarrow 59 a^{2}+27 b^{2}+9 c^{2} \leq 95$$

This is because $a \leq 1$. Equality holds if and only if $a=b=c$.