Maximum of $\frac{1}{a^2-4a+9}+\frac{1}{b^2-4b+9}+\frac{1}{c^2-4c+9}$ for $a, b, c \ge 0$; $a+b+c=1$ 
If $a+b+c=1$ and $a, b, c\geq0$, 
what is the maximum value of $\frac{1}{a^2-4a+9}+\frac{1}{b^2-4b+9}+\frac{1}{c^2-4c+9}$  ?

I found its answer by using CAS Calculator. The answer is $\frac{7}{18} (a=b=0, c=1)$. 
But i wonder not only the answer but also the solution of it.
Please help me.
 A: We have, for all $x\in [0, 1]$,
$$\frac{1}{18}x + \frac{1}{9} - \frac{1}{x^2 - 4x + 9}
= \frac{x(x-1)^2}{18(x^2-4x+9)} \ge 0. \tag{1}$$
Using (1), we have
$$\frac{1}{a^2 - 4a + 9}
+ \frac{1}{b^2 - 4b + 9} + \frac{1}{c^2 - 4c + 9}
\le \frac{1}{18}(a + b + c) + \frac{1}{3} = \frac{7}{18}. $$
Also, when $a= b = 0, c = 1$,
we have $\frac{1}{a^2 - 4a + 9}
+ \frac{1}{b^2 - 4b + 9} + \frac{1}{c^2 - 4c + 9} = \frac{7}{18}$.
Thus, the maximum of
$\frac{1}{a^2 - 4a + 9}
+ \frac{1}{b^2 - 4b + 9} + \frac{1}{c^2 - 4c + 9}$
is $7/18$.
A: This is something of a brute force method.
Let $f(a,b,c)=\sum_{a,b,c}1/ (a^2-4a+9)$ and $g(a,b,c)=a+b+c-1$ then calculate $\nabla f-\lambda \nabla g=0$ to find
$$
{2a-4\over (a^2-4a+9)^2}-\lambda = 0
$$
and similar for $b$, $c$. On the range $[0,1]$, this is a strictly increasing function of $a$, so equating the similar equations for $b$ and $c$ gives us a critical point where $a=b=c=\frac13$. Because we can set $a=b=0, c=1$ and get a larger value $f(0,0,1)={7\over18}$ than the value of $f(\frac13,\frac13,\frac13)={27\over70}$, it means this critical point is a minimum, so the maximum is achieved on the boundary where one or two of $a, b$ and $c$ are zero.
Let $a=0$ then $b+c=1$ and we have
$$
f(0,1-c,c)={1\over9}+{1\over(c+1)^2+5}+{1\over(c-2)^2+5}
$$
By graphing or inspecting values we have that this is a maximum on $[0,1]$ when $c=0$ or $c=1$, hence the maximum is at $(0,0,1)$.
