Due to the symmetry, we may assume WLOG that $a\ge b\ge c$. Therefore, $$\sqrt{a^2+bc}+\sqrt{b^2+ca}+\sqrt{c^2+ab} \le \sqrt{a^2+ca}+\sqrt{b^2+ca}+\sqrt{bc+ab}.$$By using the Power-mean inequality and the AM-GM inequality respectively, we have \begin{align*} \sqrt{a^2+ca}+\sqrt{b^2+ca}+\sqrt{bc+ab} &\le \sqrt{a^2+ca}+\sqrt{2(b^2+ca+bc+ab)} \\&= \sqrt{a(a+c)}+\sqrt{(2b+2c)(a+b)} \\&\le \frac{a+(a+c)}{2}+\frac{(2b+2c)+(a+b)}{2} \\&= \frac{3}{2}(a+b+c).\end{align*}Combining these inequalities, we get that $$\sqrt{a^2+bc}+\sqrt{b^2+ca}+\sqrt{c^2+ab} \le \frac{3}{2}(a+b+c).$$From $a+b+c < 2$, we have $$\sqrt{a^2+bc}+\sqrt{b^2+ca}+\sqrt{c^2+ab} < 3.$$

SinaQane wrote: For positive numbers $a, b, c$ satisfying condition $a+b+c<2$, Prove that $$ \sqrt{a^2 +bc}+\sqrt{b^2 +ca}+\sqrt{c^2 + ab}<3. $$ Brazilian 2013 Let $a,b,c\geq 0. $ Prove that$$ \sqrt{a^2+bc}+\sqrt{b^2+ca}+\sqrt{c^2+ab} \le \frac{3}{2}(a+b+c)$$p/7192407099