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  

Let $f(x) = \dfrac{1}{x^2+(3-x)^2}$.  WLOG let $a \le b \le c$.  
We note:  $f(x) = f(3-x)$ and
if $x < 0$, then $f(x) < f(-x)$ as is obvious from signs or from $f(x)-f(-x) = \dfrac{12x}{4x^4+81}$.  Using these, if $a < 0$, we also note 
$$f(a)+f(b)+f(c) < f(-a) + f(3-b)+f(3-c)$$ 
where the new arguments also fulfil the constraint.  
Thus if $a < 0$, to prove the inequality holds for $(a, b, c)$, it is sufficient to show it holds for $(-a, 3-b, 3-c)$.  If this also has negative variables, then one more application turns $(-a, 3-b, 3-c) \to (3+a, b, c-3)$, where the lower most value has increased by $3$ and the largest has decreased by $3$. 
Using successive application of this as necessary, it is sufficient to consider cases where $a \ge 0$, i.e. where all the variables are non-negative.  Now we look at the following version of the inequality,
$$-\sum_{cyc} \left(\frac{5}{2a^2-6a+9} - 1 \right) = \sum_{cyc} \frac{2(a-1)(a-2)}{2a^2-6a+9}$$
$$\frac{2(a-1)(a-2)}{2a^2-6a+9} = -\frac{2(a-1)}{5} + \frac{2(a-1)^2(2a+1)}{5(2a^2-6a+9)} \ge -\frac{2(a-1)}{5}$$
$$\implies -\sum_{cyc} \left(\frac{5}{2a^2-6a+9} - 1 \right) \ge \sum_{cyc}-\frac{2(a-1)}{5}=0 $$
Hence proved.
