$a,b,c\ge 0$,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


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

Thank you

  • $\begingroup$ The inequality looks very Karamata-able. Failing that, you could Muirhead bash, though that is never fun. $\endgroup$ – Ayesha May 8 '14 at 2:41
  • $\begingroup$ Perhaps this helps. Call the LHS $f(a,b,c)$. Because $f$ is homogeneous, you can assume $b=1$. Do so, and let $a=1+x$ and $b=1+y$. Now $f(a,b,c)=f(1+x,1,1+y)=\dfrac{x+1}{11 x+y+21}+\dfrac{1}{x+9 y+21}+\dfrac{y+1}{9 x+11 y+21}$. Call this function $g(x,y)$. The original problem is now equivalent to showing that when $x\ge-1$ and $y\ge-1$, $f(x,y)$ attains its maximum when $x=y=0$. This may be simpler to handle than the original question. $\endgroup$ – Steve Kass May 8 '14 at 5:03
  • $\begingroup$ You may want to watch math.stackexchange.com/questions/788988/… also. $\endgroup$ – Macavity May 11 '14 at 3:59
  • $\begingroup$ My guess is that the 7 comes from (11+9+1)/3. $\endgroup$ – marty cohen May 11 '14 at 4:32

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 $t \in \mathbb{R}_+$.

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.

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  • $\begingroup$ The reference for Theorem 1.1 is this paper: gil.ro/downloadable/download/sample/sample_id/1 , so the proof my be there. $\endgroup$ – Steve Kass May 11 '14 at 5:07
  • $\begingroup$ @SteveKass Thanks. However in that excerpt, the outline of the proof is given, referring to some "principle of global derivative" which seems to be elsewhere in the book. I will look for a copy of the book somewhere. Also the problem above is solved as well in the article, using the same "CD3 theorem". $\endgroup$ – Macavity May 11 '14 at 5:24

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!

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