Solution:

By squaring both sides and using the equality $1-\left(a^{2}+b^{2}+c^{2}\right)=2(ab+bc+ca)$, the inequality from the problem reduces to
$$\begin{aligned} L= & 2 a^{2} b+2 b^{2} c+2 c^{2} a+ \\ & 2 a b \sqrt{(2 b+1)(2 c+1)}+2 b c \sqrt{(2 c+1)(2 a+1)}+2 c a \sqrt{(2 a+1)(2 b+1)} \\ \leqslant D= & 4(ab+bc+ca) . \end{aligned}$$
By the AM-GM inequality we have $2ab \sqrt{(2b+1)(2c+1)} \leqslant ab(2b+2c+2)$ and analogously $2bc \sqrt{(2c+1)(2a+1)} \leqslant bc(2c+2a+2)$ and $2ca \sqrt{(2a+1)(2b+1)} \leqslant ca(2a+2b+2)$, so by adding we obtain
$$\begin{aligned} L \leqslant & 2\left(a^{2} b+b^{2} c+c^{2} a+ab^{2}+bc^{2}+ca^{2}+3abc\right)+2(ab+bc+ca)= \\ & 2(a+b+c+1)(ab+bc+ca)=4(ab+bc+ca)=D \end{aligned}$$

Second solution. The function $f(x)=\sqrt{x}$ is concave because $f^{\prime}(x)=2 / \sqrt{x}$ is a decreasing function. By applying Jensen's inequality with weights $a, b$ and $c$ we obtain
$$\begin{aligned} a \sqrt{2 b+1}+b \sqrt{2 c+1}+c \sqrt{2 a+1} & \leqslant \sqrt{a(2 b+1)+b(2 c+1)+c(2 a+1)} \\ & =\sqrt{1+2(ab+bc+ca)}=\sqrt{2-\left(a^{2}+b^{2}+c^{2}\right)} \end{aligned}$$