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58 posts

cadiTM

#1 h Aug 28, 2012, 1:50 PM • 6 Y

Y by yassinelbk007, Davi-8191, HWenslawski, Adventure10, Mango247, and 1 other user

Find the maximal value of the following expression, if $a,b,c$ are nonnegative and $a+b+c=1$ .
$\frac{1}{a^2 -4a+9} + \frac {1}{b^2 -4b+9} + \frac{1}{c^2 -4c+9}$

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Dr Sonnhard Graubner

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Dr Sonnhard Graubner

#2 h Aug 28, 2012, 3:12 PM • 4 Y

Y by HWenslawski, Adventure10, Mango247, and 1 other user

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.

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202 posts

#3 h Aug 30, 2012, 10:25 AM • 7 Y

Y by Ikeronalio, Andrax, Mathematicalx, aidan0626, Adventure10, Mango247, and 1 other user

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
$(a,b,c) = (1,0,0),(0,1,0),(0,0,1)$

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5 posts

Andrax

#4 h Apr 22, 2013, 12:34 PM • 2 Y

Y by Adventure10, Mango247

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?

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27 posts

#5 h Apr 22, 2013, 3:44 PM • 3 Y

Y by Andrax, Adventure10, Mango247

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 $a\ge 0$

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Andrax

#6 h Apr 23, 2013, 8:49 PM • 1 Y

Y by Adventure10

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

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6872 posts

#7 h Apr 25, 2013, 2:14 AM • 8 Y

Y by programjames1, ayan_mathematics_king, HappyMathEducation, Vladimir_Djurica, ehuseyinyigit, Adventure10, Mango247, and 1 other user

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.

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15849 posts

xzlbq

#8 h May 19, 2013, 3:45 AM • 2 Y

Y by Adventure10, Mango247

Let $a,b,c>0,a+b+c=1$ ,prove that:

$\frac{1}{a^2-4a+9}+\frac{1}{b^2-4b+9}+\frac{1}{c^2-4c+9} \geq \frac{81}{245}+\frac{27}{245}\sum_{cyc}{ac}+\frac{27}{490}\sum_{cyc}{a^2}$

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15849 posts

xzlbq

#9 h May 19, 2013, 3:54 AM • 2 Y

Y by Adventure10, Mango247

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

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15849 posts

xzlbq

#10 h May 19, 2013, 3:59 AM • 2 Y

Y by Adventure10, Mango247

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

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15849 posts

xzlbq

#11 h May 19, 2013, 4:37 AM • 2 Y

Y by Adventure10, Mango247

Let $a,b,c>0,a+b+c=1$ ,prove that:

$\frac{1}{a^2-4a+9} \geq \frac{27}{490}+\frac{27}{490}c+\frac{27}{245}a+\frac{27}{490}b$

=>
$\frac{1}{a^2-4a+9}+\frac{1}{b^2-4b+9}+\frac{1}{c^2-4c+9}\geq \frac{27}{70}$

Click to reveal hidden text

BQ

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15849 posts

xzlbq

#12 h May 19, 2013, 4:43 AM • 2 Y

Y by Adventure10, Mango247

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

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1512 posts

mathuz

#13 h May 19, 2013, 5:24 AM • 1 Y

Y by Adventure10

very beautiful solution!
I am glad.

sky0526

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308 posts

#15 h Oct 17, 2020, 3:06 AM

Y by

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.

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11199 posts

#16 h Jul 30, 2021, 6:00 PM

Y by

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 $(a,b,c)=(1,0,0)$ .

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309 posts

#17 h Jan 6, 2022, 3:14 PM

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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.

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41583 posts

sqing

#18 h Jan 7, 2022, 12:47 AM

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

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5001 posts

#20 h Oct 19, 2022, 9:02 PM • 1 Y

Y by Mango247

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. $\blacksquare$

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8857 posts

#21 h Mar 19, 2023, 6:01 PM

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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 $(1, 0, 0)$ .

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41583 posts

sqing

#22 h Mar 20, 2023, 8:38 AM

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

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Taco12

#23 h Apr 1, 2023, 12:53 AM

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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 $(1,0,0)$ .

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eibc

#24 h Apr 18, 2023, 1:08 AM

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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.

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#25 h Feb 27, 2024, 3:52 AM

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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 $a \in [0, 1]$ $\blacksquare$

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