How prove this $\frac{1}{2a^2-6a+9}+\frac{1}{2b^2-6b+9}+\frac{1}{2c^2-6c+9}\le\frac{3} {5}\cdots (1)$ let $a,b,c$ are real numbers,and such $a+b+c=3$,show that
$$\dfrac{1}{2a^2-6a+9}+\dfrac{1}{2b^2-6b+9}+\dfrac{1}{2c^2-6c+9}\le\dfrac{3}
{5}\cdots (1)$$
I find sometimes,and I find this same problem:
let $a,b,c$ are real numbers,and such $a+b+c=3$,show that
$$ \frac{1}{5a^2-4a+11}+\frac{1}{5b^2-4b+11}+\frac{1}{5c^2-4c+11}\leq\frac{1}{4} $$
and this problem have some methods,you can see:http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&t=223910&start=20
and I like this can_hang2007 methods and Honey_S methods,But for $(1)$ I can't prove it.Thank you  

A comment to complete the other  proof.

We try to show that it is enough to consider the inequality for only positive $a,b,c$. Assume $a\leq b\leq c$.
$$
\sum_{cyc}\dfrac{1}{2a^2-6a+9}=
\sum_{cyc}\dfrac{1}{(b+c)^2+a^2}=
\sum_{cyc}\dfrac{1}{9-2ab-2ac}
$$
Now it can be seen that if $a\leq 0$ and $b\leq 0$, one can choose $(-a,-b,c)$ instead and decrease the denominator and hence increase the whole sum (similar for the case where $a<0$).
In other words, for each triple $(a,b,c)$ for which at least one of $a,b,c$ is negative, one can find another triple with whole positive elements such that the sum is increased. 
