Let $a,b,c$ be positive numbers . Prove the following inequality:
$$\sqrt{\frac{11a}{5a+6b}}+\sqrt{\frac{11b}{5b+6c}}+\sqrt{\frac{11c}{5c+6a}} \leq 3.$$
What I tried:
I used Cauchy-Schwarz in the following form $\sqrt{Ax}+\sqrt{By}+\sqrt{Cz} \leq \sqrt{(a+b+c)(x+y+z)}$ for: $$A=11a, \quad{} B=11b, \quad{} C=11c$$ and $$x=\frac{1}{5a+6b}, \quad{} y=\frac{1}{5b+6c}, \quad{} z=\frac{1}{5c+6a}$$ but still nothing. Thanks for your help :)
I tried something else:
$$\large\frac{\sqrt{\frac{11a}{5a+6b}}+\sqrt{\frac{11b}{5b+6c}}+\sqrt{\frac{11c}{5c+6a}}}{3} \leq \sqrt{\frac{\frac{11a}{5a+6b}+\frac{11b}{5b+6c}+\frac{11c}{5c+6a}}{3}}$$ and what we have to prove become:
$$\large\sqrt{\frac{\frac{11a}{5a+6b}+\frac{11b}{5b+6c}+\frac{11c}{5c+6a}}{3}} \leq 1 \Leftrightarrow \sqrt{\frac{11a}{5a+6b}+\frac{11b}{5b+6c}+\frac{11c}{5c+6a}} \leq \sqrt{3}$$
Another attempt
$$\large\sqrt{\frac{1}{xy}} \leq \frac{\frac{1}{x}+\frac{1}{y}}{2}=\frac{x+y}{2xy}$$ and $y=1$ and $\displaystyle x=\frac{5a+6b}{11a}$. So:
$$\large\sqrt{\frac{11a}{5a+6b}} \leq \frac{\frac{5a+6b}{11a}+1}{2 \cdot \frac{5a+6b}{11a}}=\frac{8a+3b}{5a+6b}.$$ Now we have to prove: $$\sum_{cyc}{\frac{8a+3b}{5a+6b}} \leq 3.$$
But still nothing .