Inequality. $\sqrt{\frac{11a}{5a+6b}}+\sqrt{\frac{11b}{5b+6c}}+\sqrt{\frac{11c}{5c+6a}} \leq 3$ 
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 .

It's a bit ugly (actually more than a bit), but it works.
By AM-GM we get
$$
abc = \sqrt[3]{ab^2\cdot bc^2 \cdot ca^2} \leq \frac 1 3 (ab^2 + bc^2 + ca^2)
$$
and therefore
$$
9(5a + 6b)(5b + 6c)(5c + 6a) - 11^3(a + b + c)(ab + bc + ca) =\\ 289(ab^2 + bc^2 + ca^2) + 19(a^2b + b^2c + c^2a) - 924abc =\\ 289(ab^2 + bc^2 + ca^2 - 3abc) + 19(a^2b + b^2c + c^2a - 3abc) \geq 0
$$
From the previous inequality and applying Cauchy-Schwarz we arrive to
$$
\sum_{cyc}\sqrt{\frac {11(5c + 6a)} {(5a + 6b)(5b + 6c)(5c + 6a)} \cdot a(5b + 6c)} \leq\\
 \sqrt{\frac {11^2 (a + b+ c)} {(5a + 6b)(5b + 6c)(5c + 6a)} \sum_{cyc} a(5b + 6c) }  =\\
 \sqrt{\frac {11^3 (a + b+ c) (ab + bc + ca)} {(5a + 6b)(5b + 6c)(5c + 6a)}} \leq \sqrt 9 = 3
$$
