<s>[INST] 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.
 [/INST] 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 $7/18$.
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