hello, with the help of the Lagrange Multiplier method we get $\frac{7}{18}$ as the searched maximum, which will be attained for $(1;0;0);(0;1;0),(0;0;1)$ Sonnhard.

Since \[ \frac{1}{a^2 - 4a + 9 } \le \frac {a+2}{18} \] We get \[ \frac{1}{a^2 -4a+9} + \frac {1}{b^2 -4b+9} + \frac{1}{c^2 -4c+9} \le \frac{a+b+c+6}{18} = \frac{7}{18} \] The equality holds when \[ (a,b,c) = (1,0,0),(0,1,0),(0,0,1) \]

how did you know that 1/a^2-4a+9 < a+2/18 i'm just getting into inequalities and i don't know how you got that dosen't look like am gm hm or cauchy shwarz to me?

Andrax wrote: how did you know that 1/a^2-4a+9 < a+2/18 i'm just getting into inequalities and i don't know how you got that dosen't look like am gm hm or cauchy shwarz to me? $\frac{1}{a^2-4a+9 }\le\frac{a+2}{18}$$ \Leftrightarrow $ $ \Leftrightarrow $ $ a^3 - 4a^2 + 9a + 2a^2 - 8a +18 \ge 18$ $ \Leftrightarrow $ $ a^3 - 2a^2 + a \ge 0$ $ \Leftrightarrow $ $a(a-1)^2 \ge 0$ , which is true for $a\ge 0$

thank you very much but this seems rather odd to just bump into someone's head :/ i was rather hoping for a formula i knew i can just factor everything

Once you guess that the maximum occurs at $(1,0,0)$ you just pick the $m$ and $k$ such that $\tfrac{1}{a^2-4a+9} \le ma+k$ holds with equality at $a=0$ and $a=1$. You get lucky with $1$ being a double root though.

Let $a,b,c>0,a+b+c=1$,prove that: \[\frac{1}{a^2-4a+9}+\frac{1}{b^2-4b+9}+\frac{1}{c^2-4c+9} \geq \frac{81}{245}+\frac{27}{245}\sum_{cyc}{ac}+\frac{27}{490}\sum_{cyc}{a^2}\]

xzlbq wrote: Let $a,b,c>0,a+b+c=1$,prove that: \[\frac{1}{a^2-4a+9}+\frac{1}{b^2-4b+9}+\frac{1}{c^2-4c+9} \geq \frac{81}{245}+\frac{27}{245}\sum_{cyc}{ac}+\frac{27}{490}\sum_{cyc}{a^2}\] You do this: \[\frac{1}{a^2-4a+9} \geq \frac{27}{245}+\frac{27}{490}ac+\frac{27}{490}ab+\frac{27}{490}a^2\] in fact, \[f(a,b,c)=\frac{1}{a^2-4a+9}-\frac{27}{245}-\frac{27}{490ac}-\frac{27}{490}ab-\frac{27}{490}a^2\] \[f(\frac{x}{x+y+z},\frac{y}{x+y+z},\frac{z}{x+y+z})=\frac{1}{490}\frac{(x+4y+4z)(2x-y-z)^2}{(x+y+z)(6x^2+14xy+14xz+9y^2+18yz+9z^2)}.\] BQ

xzlbq wrote: xzlbq wrote: Let $a,b,c>0,a+b+c=1$,prove that: \[\frac{1}{a^2-4a+9}+\frac{1}{b^2-4b+9}+\frac{1}{c^2-4c+9} \geq \frac{81}{245}+\frac{27}{245}\sum_{cyc}{ac}+\frac{27}{490}\sum_{cyc}{a^2}\] You do this: \[\frac{1}{a^2-4a+9} \geq \frac{27}{245}+\frac{27}{490}ac+\frac{27}{490}ab+\frac{27}{490}a^2------------(1)\] in fact, \[f(a,b,c)=\frac{1}{a^2-4a+9}-\frac{27}{245}-\frac{27}{490ac}-\frac{27}{490}ab-\frac{27}{490}a^2\] \[f(\frac{x}{x+y+z},\frac{y}{x+y+z},\frac{z}{x+y+z})=\frac{1}{490}\frac{(x+4y+4z)(2x-y-z)^2}{(x+y+z)(6x^2+14xy+14xz+9y^2+18yz+9z^2)}.\] BQ Not setting $a->\frac{x}{x+y+z}$,you do (1)? BQ

Let $a,b,c>0,a+b+c=1$,prove that: \[\frac{1}{a^2-4a+9} \geq \frac{27}{490}+\frac{27}{490}c+\frac{27}{245}a+\frac{27}{490}b\] => \[\frac{1}{a^2-4a+9}+\frac{1}{b^2-4b+9}+\frac{1}{c^2-4c+9}\geq \frac{27}{70}\]

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BQ

Very nice is this: \[\frac{1}{x^2-4x+9}-\frac{27}{245}-\frac{27}{490}x=\frac{1}{27}\frac{(-3x+4)(-1+3x)^2}{x^2-4x+9}\] BQ

very beautiful solution! I am glad. sky0526

The answer is $\frac{7}{18}$, acheived when $(a,b,c)=(1,0,0)$. Now we show that it is the maximum. Let $f(x)=\frac{1}{x^2-4x+9}$. Suppose $a\geq b\geq c$, then $b+c\leq \frac{2}{3}$. We have $$f''(x)=\frac{3x^2-12x+7}{(x^2-4x+9)^3}$$hence it is convex in the interval $[0,\frac{2}{3}]$. Therefore, by majorization inequality. $$f(a)+f(b)+f(c)\leq f(a)+f(b+c)+f(0)$$Hence we can assume $c=0$, and try to maximize $$\frac{1}{a^2-4a+9}+\frac{1}{b^2-4b+9}=\frac{1}{a^2-4a+9}+\frac{1}{a^2+2a+6}$$WLOG assume $a\geq \frac{1}{2}$, then $$\frac{d}{da}\left(\frac{1}{a^2-4a+9}+\frac{1}{a^2+2a+6}\right)=\frac{(2a-1)(a(a-1)(a^2-a-21)+2a+17)}{(a^2-4a+9)^2(a^2+2a+6)^2}>0$$Hence we are done.

Note that $$\frac1{x^2-4x+9}\le\frac{x+2}{18}\Leftrightarrow x(x-1)^2\ge0,$$summing this for $a,b,c$ gives a maximal value of $\boxed{\frac7{18}}$ achieved at least when $(a,b,c)=(1,0,0)$.

Lets assume the function $f(x)=\frac{1}{x^2-4a+9}$. What we know is: \begin{align*} f(1)&=\frac{1}{6}\\ f'(x)&=-\frac{2(x-2)}{(x^2-4x+9)^2}\\ f'(1)&=\frac{1}{9} \end{align*}So the tangent line at $1, f(1)$ is of the form $x-18y+2=0$. Now we can say \begin{align*} f(x)\leq \frac{x+2}{18}\Longleftrightarrow x(x-1)^2\geq 0 \end{align*}which is true for all positive $x$, in particularly on the interval $[0, 1]$. Now summing them cyclically we have \begin{align*} \sum_{cyc}f(a)&\leq \sum_{cyc}\frac{a+2}{18}\\ &=\frac{6+(a+b+c)}{18}\\ &=\frac{7}{18} \end{align*}The maximum is achieved when $(a, b, c)=(1, 0, 0)$ and its permutation.

Let $a,b,c$ are nonnegative and $a+b+c=1$. Prove that $$ \frac{9}{70}\leq \frac{a}{a^2 -4a+9} + \frac {b}{b^2 -4b+9} + \frac{c}{c^2 -4c+9} \leq \frac{1}{6}$$

We have $$\frac{1}{x^2-4x+9} \leq \frac{x+2}{18} \iff (x-1)^2x\geq 0,$$which is obviously true for $x \in [0,1]$. This yields the upper bound of $\frac{1+6}{18}=\frac{7}{18}$. Since $(a,b,c)=(0,0,1)$ indeed achieves this maximum, we are done. $\blacksquare$

Notice that $$\frac 1{a^2-4a+9} \leq \frac a{18} + \frac 19 \iff \frac{x(x-1)^2}{18(x^2-4x+9)} \geq 0,$$which is evident. Thus, summing cyclically yields a minimum of $\frac 7{18}$, with equality at $(1, 0, 0)$.

Let $a,b,c$ are nonnegative and $a+b+c=1$. Prove that $$ \frac{11}{18}\leq \frac{2}{a^2 -4a+9} + \frac {1}{b^2 -4b+9} + \frac{2}{c^2 -4c+9} \leq \frac{2}{3}$$$$ \frac{130}{261}\leq \frac{1}{a^2 -4a+9} + \frac {2}{b^2 -4b+9} + \frac{1}{c^2 -4c+9} \leq \frac{5}{9}$$

Let $f(x)=\frac{1}{x^2 - 4x + 9 }$. We claim $f(x) \le \frac {x+2}{18},$ which is obviously true. Then, summing $f(a), f(b), f(c)$ yields $$f(a)+f(b)+f(c)\leq \frac{a+b+c+6}{18}=\frac{7}{18}.$$Equality occurs at $(1,0,0)$.

From the tangent line trick, we can prove that $$\frac{1}{x^2 - 4x + 9} \le \frac{x}{18} + \frac{1}{9} \iff (x - 1)^2x \ge 0,$$which is clearly true for any $0 \le x \le 1$. Hence summing cyclically in $a$, $b$, and $c$, we find that the expression in question reaches a maximum of $\tfrac{7}{18}$ at $(a, b, c) = (1, 0, 0)$ and permutations.

We claim that the maximum value is $\boxed{\frac{7}{18}}$. This is given by $(a, b, c) = (1, 0, 0)$ and permutations. Now we prove that this is the maximum. We claim that \[\frac{1}{a^2-4a+9} \leq \frac{a+2}{18}\]Where $a \in [0, 1]$. Obviously this proves the problem. Now, we multiply both sides by $18a^2-72a+162$. This gives \[a^3-2a^2+a \geq 0 \Rightarrow a(a-1)^2 \geq 0\]Which is obviously true for $a \in [0, 1]$ $\blacksquare$