Proof of an inequality

If $a$, $b$, $c$ are positive real numbers, prove that $$\frac{\sqrt{a+b+c}+\sqrt{a}}{b+c} + \frac{\sqrt{a+b+c}+\sqrt{b}}{c+a} + \frac{\sqrt{a+b+c}+\sqrt{c}}{a+b} \geq \frac{9+3\sqrt{3}}{2\sqrt{a+b+c}}$$

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What are your thoughts? What have you tried and where are you getting stuck? – vociferous_rutabaga Jul 23 '14 at 3:28
Note $$b+c = (a+b+c)-a = (\sqrt{a+b+c}-\sqrt{a})(\sqrt{a+b+c}+\sqrt a)$$ – Thomas Andrews Jul 23 '14 at 3:32
To be humble, @MorganO, I tried several things - but finally none of them seemed to be on a correct direction. The last one I tried is multiplying both sides by $\sqrt{a+b+c}$. – pushpen.paul Jul 23 '14 at 3:38

As the inequality is homogeneous, we can set $a+b+c = 1$, say. Then we have to show the cyclic sum: $$\sum_{cyc} \frac{1+\sqrt{a}}{1-a} = \sum_{cyc} \frac1{1- \sqrt a} \ge \frac{9+3\sqrt3}2$$

To show this, it is sufficient to show that $$f(x) = \frac1{1-\sqrt x} - \frac{3+\sqrt 3}2 - k(\tfrac13-x) \ge 0$$ for some $k \in \mathbb R$ and $x \in (0, 1)$, as the inequality is equivalent to $f(a)+f(b)+f(c) \ge 0$.

We find that $k = \frac34(3+2\sqrt 3)$ works, as then $$f(x) = \frac{(3 + 2 \sqrt3) \left(\sqrt3 - 3 \sqrt x \right)^2 (-3 + 2 \sqrt3 + 3 \sqrt x)}{36(1 - \sqrt x)} \ge 0, \quad \forall x \in (0, 1)$$

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Had to change my answer as $t \mapsto \dfrac1{1-\sqrt t}$ is not convex close to $0$, so Jensen cannot be directly applied. – Macavity Jul 23 '14 at 5:15
Perhaps easier would be Cauchy-Schwarz: with $$3 = (a+b+c)(1+1+1) \ge \left(\sum_{cyc} \sqrt a\right)^2 \implies \sum_{cyc} (1-\sqrt a) \le \frac6{3+\sqrt3}$$ and $$\left( \sum_{cyc} \frac1{1- \sqrt a} \right) \left(\sum_{cyc} (1-\sqrt a) \right) \ge 9$$ – Macavity Jul 23 '14 at 5:27