Please help with this hard inequality provement Please help to prove this.
Assume $a,b,c,m,n>0$, then we have
$${a\over ma+nb+c}+{b \over  {mb+nc+a}}+{c\over{mc+na+b}}\leq\max\{{3\over m+n+1},{1\over m},{2\over m+\sqrt{n}}\}$$
Thanks.
 A: Haven't found a simple method, but on clearing denominators, this would be a third degree cyclic homogeneous polynomial inequality, so the Theorem 1.1 is applicable.
Let $\displaystyle f(a, b, c) = \sum_{cyc} \frac{a}{ma+nb+c}$.  To find the maximum of $f$, we need equivalently to find the maximum of $f(1, 1, 1)$ and $f(a, 1, 0)$.  The first is obviously $\dfrac3{m+n+1}$, and for the second, we need the maximum of 
$$g(a) = \frac{a}{ma+n}+\frac{1}{m+a}$$
The simplest way now is by derivatives, and we get $\displaystyle g'(a) = -\frac{(a^2-n)(m^2-n)}{(a+m)^2(am+n)^2}$.
So $g'(a)=0$ has one positive root, $a = \sqrt n$.  
From the expression for $g'(a)$ it is clear that if $m^2 < n$, $g$ is increasing beyond that root, so the maximum is $\lim_{a \to \infty} g(a) = \dfrac1m$.  OTOH, if $m^2 > n$, then the maximum is $g(\sqrt n) = \dfrac2{m+\sqrt n}$.
Putting all that together, 
$$f(a) \le \max \left(\frac3{m+n+1}, \frac1m, \frac2{m+\sqrt n}\right)$$
Equality is either when $a=b=c$, or when $m^2 > n$ and $(a, b, c)$ is a cyclic permutation of $(\sqrt n \,t, t, 0)$ for any $t \in \mathbb{R}_+$.

N.B.  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.
