Proof of an inequality

Question

Let $a$, $b$ and $c$ be positive numbers. Prove that: $$\frac{\sqrt{a+b+c}+\sqrt{a}}{b+c} + \frac{\sqrt{a+b+c}+\sqrt{b}}{c+a} + \frac{\sqrt{a+b+c}+\sqrt{c}}{a+b} \geq \frac{9+3\sqrt{3}}{2\sqrt{a+b+c}}$$

Answer 1

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 $$f(x) = \frac{(3 + 2 \sqrt3) \left(\sqrt3 - 3 \sqrt x \right)^2 (-3 + 2 \sqrt3 + 3 \sqrt x)}{36(1 - \sqrt x)} \ge 0, \quad \forall x \in (0, 1)$$

Answer 2

We need to prove that $$\sum_{cyc}\frac{a+b+c+\sqrt{a(a+b+c)}}{b+c}\geq\frac{9+3\sqrt3}{2}$$ or $$\sum_{cyc}\frac{a}{b+c}+\sum_{cyc}\frac{\sqrt{a(a+b+c)}}{b+c}\geq\frac{3+3\sqrt3}{2}.$$ But by C-S $$\sum_{cyc}\frac{a}{b+c}=\sum_{cyc}\frac{a^2}{ab+ac}\geq\frac{(a+b+c)^2}{2(ab+ac+bc)}\geq\frac{3}{2}.$$ Thus, it remains to prove that $$\sum_{cyc}\frac{\sqrt{a}}{b+c}\geq\frac{3\sqrt{3}}{2\sqrt{a+b+c}}.$$ Now, by Holder $$\left(\sum_{cyc}\frac{\sqrt{a}}{b+c}\right)^2\sum_{cyc}a^2(b+c)^2\geq(a+b+c)^3.$$ Id est, it remains to prove that $$4(a+b+c)^4\geq27\sum_{cyc}a^2(b+c)^2,$$ which is true by AM-GM: $$27\sum_{cyc}a^2(b+c)^2=108\sum_{cyc}a\left(a\left(\frac{b+c}{2}\right)^2\right)\leq$$ $$\leq108\sum_{cyc}a\left(\frac{a+\frac{b+c}{2}+\frac{b+c}{2}}{3}\right)^3=4\sum_{cyc}a(a+b+c)^3=4(a+b+c)^4.$$ Done!