Problem 2091

We first observe that $a^2 + 1 \ge a^2 + ab + bc + ca \ge 4a\sqrt{bc}$ where the second inequality results from $A.M. \ge G.M.$. Therefore we have $2\sqrt{bc} \ge \frac{8abc}{a^2+1}$. Summing this up with similar inequalities for $b$ and $c$ gives that it suffices to show that
$$a + b + c + \sqrt{3} \ge 2(\sqrt{ab} + \sqrt{bc} + \sqrt{ca}).$$
By the Cauchy-Schwarz inequality and $1 \ge ab + bc + ca$, we have
$$\sqrt{3} \ge \sqrt{1+1+1\sqrt{ab+bc+ca}} \ge \sqrt{ab} + \sqrt{bc} + \sqrt{ca}.$$
As $(\sqrt{a} - \sqrt{b})^2, (\sqrt{b} - \sqrt{c})^2, (\sqrt{c} - \sqrt{a})^2 \ge 0$ we obtain
$$ab + bc + ca \ge \sqrt{ab} + \sqrt{bc} + \sqrt{ca}$$
and the result follows.