How prove $\frac{a}{11a+9b+c}+\frac{b}{11b+9c+a}+\frac{c}{11c+9a+b}\le\frac{1}{7}$ Qustion:

$a,b,c\ge 0$,show that
  $$\dfrac{a}{11a+9b+c}+\dfrac{b}{11b+9c+a}+\dfrac{c}{11c+9a+b}\le\dfrac{1}{7}\tag{1}$$

I found this method can't work,
$$x=11a+9b+c,y=11b+9c+a,z=11c+9a+b$$
$$\Longrightarrow a=\dfrac{8x-7y+5z}{126},b=\dfrac{5x+8y-7z}{126},c=\dfrac{5y-7x+8z}{126}$$
so
$$LHS=\dfrac{1}{126}\left(\dfrac{8x-7y+5z}{x}+\dfrac{5x+8y-7z}{y}+\dfrac{5y-7x+8z}{z}\right)$$
so
$$LHS=\dfrac{1}{126}\left(24-7\left(\dfrac{y}{x}+\dfrac{z}{y}+\dfrac{x}{z}\right)+5\left(\dfrac{z}{x}+\dfrac{x}{y}+\dfrac{y}{z}\right)\right)$$
then I can't use AM-GM inequality to Continue,
By the way
I use AM-GM inequality can solve this  famous
 Crux problem (2009)
let $x,y,z\ge 0$,and $a,b,c>0$, if $b^2\le ca ,c^2\le ab$,then we can use this methods to solve this Crux problem

$$\dfrac{x}{ax+by+cz}+\dfrac{y}{ay+bz+cz}+\dfrac{z}{az+bx+cy}\le\dfrac{3}{a+b+c}$$

so how can we prove this (1) inequality? and I Think this inequality is very sharp,maybe there are other methods.
Thank you

Let $a=\min\{a,b,c\}$,$b=a+u$ and $c=a+v$.
Hence, we need to prove that 
$$49(u^2-uv+v^2)a+(9u+v)(u-3v)^2\geq0$$
Done! 
