<s>[INST] $a,b,c>0 \text{ s.t. }a+b+c=1 \implies \sqrt{ab+c}+\sqrt{bc+a}+\sqrt{ca+b} \ge 1+ \sqrt{ab}+\sqrt{bc}+\sqrt{ca}$ How can we show that the assumption $a,b,c>0$ and $a+b+c=1$ implies 
$$\sqrt{ab+c}+\sqrt{bc+a}+\sqrt{ca+b} \ge 1+ \sqrt{ab}+\sqrt{bc}+\sqrt{ca}~?$$
 [/INST] First we claim that $\sqrt{ab+c} \geq \sqrt{ab}+c$. 
Proof of the claim:
$$\begin{eqnarray*}
\sqrt{ab}+c 
&=& \sqrt{ab + 2c\sqrt{ab} + c^2}
\\ &\leq& \sqrt{ab + (a+b) c + c^2}
\\ &=& \sqrt{ab + c(a+b+c)} = \sqrt{ab+c}.
\end{eqnarray*}
$$
The only inequality used here is the AM-GM inequality: $\sqrt{ab} \leq \frac{a+b}{2}$. $\Box$
Now, the full inequality follows easily from the claim. By symmetry, we also have $\sqrt{bc+a} \geq \sqrt{bc}+a$ and $\sqrt{ac+b}\geq \sqrt{ac}+b$. Adding all three inequalities, we get the result. 
</s>