Lemma: $\frac{1}{x^2-4x+9} \le \frac{x+2}{18}$, since it comes to $\frac{x(x-1)^2}{x^2-4x+9} \geq 0$. Therefore: $\sum \frac{1}{x^2-4x+9} \le \sum \frac{x+2}{18} = \frac {7} {18}$. Equality holds when $x,y,z$ are $0,0,1$ in any order.

Aha got it It goes very simple this way Thnx guys !

See also here: http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&t=496112

\[\frac{1}{a^2-4a+9}+\frac{1}{b^2-4b+9}+\frac{1}{c^2-4c+9} \geq \frac{81}{245}+\frac{27}{245}\sum_{cyc}{ac}+\frac{27}{490}\sum_{cyc}{a^2}\][/quote]

xzlbq wrote: Let $a,b,c>0,a+b+c=1$,prove that: \[\frac{1}{a^2-4a+9}+\frac{1}{b^2-4b+9}+\frac{1}{c^2-4c+9} \geq \frac{81}{245}+\frac{27}{245}\sum_{cyc}{ac}+\frac{27}{490}\sum_{cyc}{a^2}\] You do this: \[\frac{1}{a^2-4a+9} \geq \frac{27}{245}+\frac{27}{490}ac+\frac{27}{490}ab+\frac{27}{490}a^2\] in fact, \[f(a,b,c)=\frac{1}{a^2-4a+9}-\frac{27}{245}-\frac{27}{490ac}-\frac{27}{490}ab-\frac{27}{490}a^2\] \[f(\frac{x}{x+y+z},\frac{y}{x+y+z},\frac{z}{x+y+z})=\frac{1}{490}\frac{(x+4y+4z)(2x-y-z)^2}{(x+y+z)(6x^2+14xy+14xz+9y^2+18yz+9z^2)}.\] BQ

xzlbq wrote: xzlbq wrote: Let $a,b,c>0,a+b+c=1$,prove that: \[\frac{1}{a^2-4a+9}+\frac{1}{b^2-4b+9}+\frac{1}{c^2-4c+9} \geq \frac{81}{245}+\frac{27}{245}\sum_{cyc}{ac}+\frac{27}{490}\sum_{cyc}{a^2}\] You do this: \[\frac{1}{a^2-4a+9} \geq \frac{27}{245}+\frac{27}{490}ac+\frac{27}{490}ab+\frac{27}{490}a^2------------(1)\] in fact, \[f(a,b,c)=\frac{1}{a^2-4a+9}-\frac{27}{245}-\frac{27}{490ac}-\frac{27}{490}ab-\frac{27}{490}a^2\] \[f(\frac{x}{x+y+z},\frac{y}{x+y+z},\frac{z}{x+y+z})=\frac{1}{490}\frac{(x+4y+4z)(2x-y-z)^2}{(x+y+z)(6x^2+14xy+14xz+9y^2+18yz+9z^2)}.\] BQ Not setting $a->\frac{x}{x+y+z}$,you do (1)? BQ

My solution: let $f(x)=\dfrac{1}{x^2-4x+9}$. By finding second derivate, we can see that $f(x)$ on $[0,1]$ is convex. Applying Petrovic's inequality we have: $f(x)+f(y)+f(z)\leq f(x+y+z) + (3-1)\cdot f(0)$. So, the expression is greater or equal than $f(1)+2f(0)=\dfrac{7}{18}$ Checking with $0,0,1$ we can see that this value can be satisfied. Petrovic's inequality: http://web.math.pmf.unizg.hr/glasnik/18.1/18110.pdf

@above (sorry for being too late) when does the equality hold in petrovic's inequality?