First of all let us show that $\sqrt{a+b} + \sqrt{b+c} + \sqrt{c+a} \ge \sqrt{7(a+b+c)-3}$ (1). Let $a+b+c = x$, then $ab+bc+ca = \frac{x^2-1}{2}$ and since by Cauchy-Schwarz inequality $1 = a^2+b^2+c^2 \le (a+b+c)^2 \le 3(a^2+b^2+c^2) = 3$ we get that $1 \le x \le \sqrt{3}$. It can be readily shown that

$$(\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a})^2 = 2x+2\left(\sqrt{a^2+\frac{x^2-1}{2}}+\sqrt{b^2+\frac{x^2-1}{2}}+\sqrt{c^2+\frac{x^2-1}{2}}\right).$$

Note that for $0 \le a \le 1$ and $x \ge 1$ $\sqrt{a^2 + \frac{x^2-1}{2}} \ge a + \frac{x-1}{2}$ (2). Indeed, by taking square of both sides we get $a^2 + \frac{x^2-1}{2} \ge a^2 + a(x-1) + \frac{x^2-1}{4}$ which is equivalent to $(x-1)(x+3-4a) \ge 0$. Since $x \ge 1 \ge a$ (2) is proved. Now (2) implies (1).

In order to complete solution let us show that $7(a+b+c) - 3 \ge (2+5abc)^2$. By AM-GM inequality $ab + bc + ca \ge 3\sqrt[3]{a^2b^2c^2}$ and therefore $(2+5abc)^2 \le (2+5(\frac{x^2-1}{6})^{3/2})^2$. Thus, it is sufficient to show that $(2+5(\frac{x^2-1}{6})^{3/2})^2 \le 7x-3$. Now

$$\begin{align*} \left(2 + 5\left(\frac{x^2-1}{6}\right)^{3/2}\right)^2 &\le 7x-3 \\ \iff 7(x-1) \ge 25\left(\frac{x^2-1}{6}\right)^3 + 20\left(\frac{x^2-1}{6}\right)^{3/2} \\ &\iff \left(\frac{25(x^2-1)^2(x+1)}{216} + \frac{5\sqrt{6}(x^2-1)^{1/2}(x+1)}{9} - 7\right)(x-1) \le 0 \end{align*}$$

Let us show that if $1 \le x \le \sqrt{3}$ then $\frac{25(x^2-1)^2(x+1)}{216} + \frac{5\sqrt{6}(x^2-1)^{1/2}(x+1)}{9} \le 7$. Since the last expression is an increasing function of $x$ for $x \ge 1$ we will check it only for $x = \sqrt{3}$ : $\frac{205+85\sqrt{3}}{54} \approx 6.52 < 7$. Done.