<s>[INST] 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  
 [/INST] We need to prove that
$$\sum_{cyc}\frac{1}{2a^2-6a+9}\leq\frac{3}{5}$$ or
$$\sum_{cyc}\left(\frac{1}{5}-\frac{1}{2a^2-6a+9}\right)\geq0$$ or
$$\sum_{cyc}\frac{a^2-3a+2}{2a^2-6a+9}\geq0$$ or
$$\sum_{cyc}\left(\frac{18(a^2-3a+2)}{2a^2-6a+9}+1\right)\geq3$$ or
$$\sum_{cyc}\frac{(2a-3)^2}{2a^2-6a+9}\geq\frac{3}{5}.$$
Now, by C-S we obtain:
$$\sum_{cyc}\frac{(2a-3)^2}{2a^2-6a+9}=\sum_{cyc}\frac{(2a-3)^2(a-8)^2}{(2a^2-6a+9)(a-8)^2}\geq$$
$$\geq\frac{\left(\sum\limits_{cyc}(2a-3)(a-8)\right)^2}{\sum\limits_{cyc}(2a-3)^2(a-8)^2}=\frac{\left(\sum\limits_{cyc}(2a^2+5)\right)^2}{\sum\limits_{cyc}(2a-3)^2(a-8)^2}.$$
Hence, it remains to prove that
$$\frac{\left(\sum\limits_{cyc}(2a^2+5)\right)^2}{\sum\limits_{cyc}(2a-3)^2(a-8)^2}\geq\frac{3}{5}.$$
Let $a+b+c=3u$, $ab+ac+bc=3v^2$, where $v^2$ can be negative, and $abc=w^3$.
Hence, we need to prove a fourth degree polynomial inequality, id est, a linear inequality of $w^3$,
which says that it's enough to prove the last inequality for an extremal value of $w^3$,
which happens for equality case of two variables.
Let $b=a$ and $c=3-2a$.
We need to prove that $(a-1)^2(34a^2-98a+73)\geq0$, which is obvious.
Done!
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