<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] Let $f(x) = \dfrac{1}{2x^2-6x+9}$. First let us first quickly sketch some properties of $f$ that we'll need. $f(x)=\dfrac{1}{2(x-\frac32)^2+\frac92}$, so its maximum is $f(\frac32)=\frac29$. $f~'(x)=\dfrac{6-4x}{(2x^2-6x+9)^2}$, and $f~''(x)=12\dfrac{2x^2-6x+3}{(2x^2-6x+9)^3}$. Setting $f~''(x)=0$ gives the inflection points $x=\frac32\pm\frac{\sqrt3}2$. $f$ is increasing before $\frac32$, decreasing after $\frac32$, and is concave between the inflection points. Note $f(\frac32\pm\frac{\sqrt3}2)=\frac16$.
If $a$, $b$, and $c$ are all between the inflection points, then by Jensen's inequality
$$f(a)+f(b)+f(c) \le 3f(\dfrac{a+b+c}3) = 3f(1) = \frac35$$
as required.
If at least two of $a$, $b$, or $c$ are outside the inflection points, then
$$f(a)+f(b)+f(c) \le \frac16+\frac16+\frac29=\frac59\lt\frac35$$
If one of $a$, $b$, or $c$ is $\gt\frac32+\frac{\sqrt3}2$, then at least one of $a$, $b$, or $c$ is $\lt\frac32-\frac{\sqrt3}2$, otherwise
$$3=a+b+c\ge 2(\frac32-\frac{\sqrt3}2)+(\frac32+\frac{\sqrt3}2)=\frac92-\frac{\sqrt3}2 \approx 3.7,$$
a contradiction. And so by the previous paragraph, the required result holds.
So the only remaining case is when exactly one of $a$, $b$, or $c$ is $\lt\frac32-\frac{\sqrt3}2$, and the other two are between the inflection points. We need two estimates. First consider the case when the minimum is $\le\frac12$. (Here $\frac12$ is arbitrary, although we need to choose something very close to $\frac12$ for the following argument to work.) Then
$$f(a)+f(b)+f(c) \le f(\frac12) + \frac29 + \frac29 = \frac{2}{13} + \frac49 = \frac{70}{117} \lt \frac35$$
Finally, suppose $a$ is between $\frac12$ and $\frac32-\frac{\sqrt3}2$, and $b$ and $c$ are between the inflection points. By Jensen's inequality, we need only consider the case when $b=c$, by replacing them with their mean. Then $b=(3-a)/2$. So $a\gt\frac12$ implies $b\lt\frac54$, and
$$f(a)+f(b)+f(b) \lt f(\frac32-\frac{\sqrt3}2) + 2f(\frac54) = \frac16 + \frac{16}{37} = \frac{133}{222} \lt \frac35$$
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