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  

Here's what I've got so far. 
It's not a complete solution,
but the ideas might be useful
to someone else.
$2x^2-6x+9
=x^2+(x-3)^2
$.
If
$f(x)
=2x^2-6x+9
$,
$f'(x)
=4x-6
$
is zero at $x = 3/2$.
Since $f(3/2)
=9/2
$,
$f(x) \ge 9/2$
for all $x$.
This means that
$\dfrac1{f(x)}
\le \dfrac{2}{9}
$
for all $x$,
so
$\dfrac1{f(a)}+\dfrac1{f(b)}+\dfrac1{f(c)}
\le \dfrac{6}{9}
= \dfrac23
$
without any hypothesis on $a, b, c$.
If $c = 3-a-b$,
$f(c)
=(3-a-b)^2+(a+b)^2
$.
If $a=b=3/2$,
$c = 0$
so
$f(c) = 9$
so
$\dfrac1{f(a)}+\dfrac1{f(b)}+\dfrac1{f(c)}
= \dfrac{5}{9}
< \dfrac{3}{5}
$.
If $a=b=c$,
$a=b=c=1$.
Since $f(1)
=1+4=5
$,
$\dfrac1{f(a)}+\dfrac1{f(b)}+\dfrac1{f(c)}
= \dfrac{3}{5}
$,
so this is the equality case.
If we can show that
$\dfrac1{f(a)}+\dfrac1{f(b)}+\dfrac1{f(3-a-b)}
$
is a maximum when
$a=b=1$,
that would do it.
But I don't know how right now,
so I'll leave it at this.
