Team Selection Test

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.