# $a,b,c>0 \text{ s.t. }a+b+c=1 \implies \sqrt{ab+c}+\sqrt{bc+a}+\sqrt{ca+b} \ge 1+ \sqrt{ab}+\sqrt{bc}+\sqrt{ca}$

How can we show that the assumption $a,b,c>0$ and $a+b+c=1$ implies

$$\sqrt{ab+c}+\sqrt{bc+a}+\sqrt{ca+b} \ge 1+ \sqrt{ab}+\sqrt{bc}+\sqrt{ca}~?$$

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@joriki : looked like a typo got edited out. –  Arjang Jul 30 '11 at 16:16

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.