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
Haven't found a simple method, but on clearing denominators, this would be a third degree cyclic homogeneous polynomial inequality, so Theorem 1.1. is applicable.
Let $\displaystyle f(a, b, c) = \frac17 - \sum_{cyc} \frac{a}{11a+9b+c}$. To show $f \ge 0$, we need equivalently to only show $f(1, 1, 1) \ge 0$ and $f(a, 1, 0) \ge 0$. The first is obvious, and the second is $$\frac17 \ge \frac{a}{11a+9}+\frac{1}{11+a} \iff (3-a)^2 \ge 0$$
Thus we have proven the inequality, and equality is when $(a, b, c) = (t, t, t)$ or a cyclic permutation of $(3t, t, 0)$ for any $t \in \mathbb{R}_+$.
NB:
1. The more general case of $\sum_{cyc} \dfrac{\alpha a + \beta b + \gamma c}{p a + q b + r c}$ can be tackled in exactly the same way.
2. I haven't seen a complete proof of this Theorem 1.1., though its quite nice. If any one has an online reference that would be great.