a+b+c=27

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KaiRain

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#1 h Oct 6, 2018, 11:04 AM • 2 Y

Y by Adventure10, cubres

Let $a,b,c$ be positive real numbers such that $a+b+c=27$ . Prove that:
$\frac{1}{a^2+155}+\frac{1}{b^2+155}+\frac{1}{c^2+155}\le \frac{11}{780}$ When does the equality hold ?

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Zark84010

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#2 h Oct 6, 2018, 11:40 AM • 3 Y

Y by Adventure10, Mango247, cubres

The function $f(x)=\dfrac{1}{155+x^{2}}$ is concave for $x>0$ (for all $x\in \mathbb{R}$ in fact). Just use Jensen's inequality.

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arqady

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#3 h Oct 6, 2018, 11:41 AM • 3 Y

Y by Adventure10, Mango247, cubres

KaiRain wrote:

Let $a,b,c$ be positive real numbers such that $a+b+c=27$ . Prove that:
$\frac{1}{a^2+155}+\frac{1}{b^2+155}+\frac{1}{c^2+155}\le \frac{11}{780}$ When does the equality hold ?

It's true for all reals $a$ , $b$ and $c$ such that $a+b+c=27$ .

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#4 h Oct 6, 2018, 11:42 AM • 3 Y

Y by Adventure10, Mango247, cubres

Zark84010 wrote:

The function $f(x)=\dfrac{1}{155+x^{2}}$ is concave for $x>0$ .

Are you sure?

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KaiRain

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#5 h Oct 6, 2018, 12:57 PM • 2 Y

Y by Adventure10, cubres

arqady wrote:

It's true for all reals $a$ , $b$ and $c$ such that $a+b+c=27$ .

Yep!

I've found general form

If $a,b,c$ are positive real numbers and $n \ge 2$ such that $a+b+c=n^2+2$ then
$\sum_{sym} \frac{1}{a^2+n(n^2+n+1)} \le \frac{2n+1}{n(n+1)(n^2+1)}$

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#7 h Oct 6, 2018, 1:09 PM • 3 Y

Y by Adventure10, Mango247, cubres

For $n=2:$
Let $a,b,c$ be positive real numbers such that $a+b+c=6.$ Prove that:
$\frac{1}{a^2+14}+\frac{1}{b^2+14}+\frac{1}{c^2+14} \le \frac{1}{6}$ When does the equality hold ?

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arqady

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#8 h Oct 6, 2018, 1:12 PM • 5 Y

Y by KaiRain, Adventure10, Mango247, ehuseyinyigit, cubres

KaiRain wrote:

When does the equality hold ?

For $(a,b,c)=(1,1,4)$ of course!

And for $(a,b,c)=(1,1,n^2)$ in the general. Nice!

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KaiRain

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#9 h Oct 6, 2018, 1:13 PM • 5 Y

Y by Adventure10, Mango247, Mango247, Mango247, cubres

arqady wrote:

KaiRain wrote:

When does the equality hold ?

For $(a,b,c)=(1,1,4)$ of course!

Nice!

Also $a=b=c=2$

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#10 h Oct 6, 2018, 2:31 PM • 4 Y

Y by KaiRain, Adventure10, Mango247, cubres

KaiRain wrote:

Also $a=b=c=2$

Yes, it's the known Can's inequality.
See here: https://artofproblemsolving.com/community/c6h238857

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cezartrancanau

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#11 h Oct 6, 2018, 2:36 PM • 2 Y

Y by Adventure10, cubres

KaiRain wrote:

If $a,b,c$ are positive real numbers and $n \ge 2$ such that $a+b+c=n^2+2$ then
$\sum_{sym} \frac{1}{a^2+n(n^2+n+1)} \le \frac{2n+1}{n(n+1)(n^2+1)}$

This is a brilliant problem.
Observe that the function $f(x)=\frac{1}{x^2+n(n^2+n+1)}$ is concave on $\left[0,\sqrt{\frac{n(n^2+n+1)}{3}}\right].$

Assume $WLOG \; a \leq b \leq c.$ If $b\leq \sqrt{\frac{n(n^2+n+1)}{3}}$ . Apply Jensen and we get
$f(a)+f(b) \leq 2f\left(\frac{a+b}{2}\right)$ Set $x=\frac{a+b}{2}.$ Then $x \in \left[0,\sqrt{\frac{n(n^2+n+1)}{3}}\right]$ and suffice it to prove
$2f(x)+f(n^2+2-2x)\leq \frac{2n+1}{n(n+1)(n^2+1)}$

We leave for the reader this wonderful part.

If $b > \sqrt{\frac{n(n^2+n+1)}{3}}$ , it is a beautiful algebraic manipulation.

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#12 h Oct 6, 2018, 3:14 PM • 3 Y

Y by Adventure10, Mango247, cubres

Thank you !

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#13 h Oct 6, 2018, 3:16 PM • 3 Y

Y by Adventure10, Mango247, cubres

cezartrancanau wrote:

Observe that the function $f(x)=\frac{1}{x^2+n(n^2+n+1)}$ is concave on $\left[0,\frac{n^2+2}{2}\right].$

Can you show more detail, since I can't get it

Thanks .

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cezartrancanau

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#14 h Oct 6, 2018, 4:14 PM • 4 Y

Y by KaiRain, Adventure10, Mango247, cubres

KaiRain wrote:

Can you show more detail, since I can't get it

Thanks .

I edited it.

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arqady

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#15 h Oct 6, 2018, 6:06 PM • 4 Y

Y by KaiRain, Adventure10, Mango247, cubres

KaiRain wrote:

Let $a,b,c$ be positive real numbers such that $a+b+c=27$ . Prove that:
$\frac{1}{a^2+155}+\frac{1}{b^2+155}+\frac{1}{c^2+155}\le \frac{11}{780}$ When does the equality hold ?

My solution.
We need to prove that
$\sum_{cyc}\frac{1}{a^2+155}\leq\frac{11}{780}$ or
$\sum_{cyc}\left(\frac{1}{a^2+155}-\frac{1}{155}\right)\leq\frac{11}{780}-\frac{3}{155}$ or
$\sum_{cyc}\frac{a^2}{a^2+155}\geq\frac{127}{156}.$ Now, by C-S
$\sum_{cyc}\frac{a^2}{a^2+155}=\sum_{cyc}\frac{a^2(a+5)^2}{(a+5)^2(a^2+155)}\geq\frac{\left(\sum\limits_{cyc}(a^2+5a)\right)^2}{\sum\limits_{cyc}(a+5)^2(a^2+155)}.$ Thus, it's enough to prove that
$156\left(\sum\limits_{cyc}(a^2+5a)\right)^2\geq127\sum\limits_{cyc}(a+5)^2(a^2+155),$ which is fourth degree.
Id est, by $uvw$ it's enough to prove the last inequality for equality case of two variables.
Let $b=a$ and $c=27-2a$ .
We obtain:
$(a-1)^2(555a^2-13028a+84449)\geq0,$ which is true even for all real value of $a$ .
The equality occurs for $(a,b,c)=(1,1,25)$ and for the cyclic permutations of this.

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sqing

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#16 h Apr 12, 2025, 2:09 PM • 1 Y

Y by cubres

Let $a,b,c$ are positive real number with $a+b+c=1.$ Prove that $\frac{1}{a^2+3} + \frac{1}{b^2+3} + \frac{1}{c^2+3} \leq \frac{27}{28}$ Let $a,b,c \geq -2$ such that $a^2+b^2+c^2 \leq 8.$ Prove that $\frac{1}{16+a^3}+\frac{1}{16+b^3}+\frac{1}{16+c^3}\leq \frac{5}{16}$ h

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SunnyEvan

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#17 h Apr 12, 2025, 2:18 PM • 1 Y

Y by cubres

sqing wrote:

Let $a,b,c$ are positive real number with $a+b+c=1.$ Prove that $\frac{1}{a^2+3} + \frac{1}{b^2+3} + \frac{1}{c^2+3} \leq \frac{27}{28}$ Let $a,b,c \geq -2$ such that $a^2+b^2+c^2 \leq 8.$ Prove that $\frac{1}{16+a^3}+\frac{1}{16+b^3}+\frac{1}{16+c^3}\leq \frac{5}{16}$ h

Let $f(x)= \frac{1}{x^2+3}$ and $x\in(0,1)$
we have : $f''(x)=\frac{6x^2-6}{x^6+9x^4+27x^2+27} <0$ use Jensen : $\frac{1}{a^2+3} + \frac{1}{b^2+3} + \frac{1}{c^2+3} =\sum f(a) \leq 3f(\frac{\sum a}{3})=\frac{27}{28}$

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SunnyEvan

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#18 h Apr 12, 2025, 2:31 PM • 1 Y

Y by cubres

sqing wrote:

Let $a,b,c$ are positive real number with $a+b+c=1.$ Prove that $\frac{1}{a^2+3} + \frac{1}{b^2+3} + \frac{1}{c^2+3} \leq \frac{27}{28}$ Let $a,b,c \geq -2$ such that $a^2+b^2+c^2 \leq 8.$ Prove that $\frac{1}{16+a^3}+\frac{1}{16+b^3}+\frac{1}{16+c^3}\leq \frac{5}{16}$ h

Let $f(x)=\frac{1}{16+x^3}$ and $x\in(-2,+\infty)$
we have $f'(x)=-\frac{3x^2}{(x^3+16)^2} <0$ meaning $f(x)$ is decreasing .Thus ,we can use H-C-F-T and $f(x)$ is maximized at the minimum value $x=2 .$
equality cases : $(a,b,c)=(-2,-2,0)$ and permutations.

This post has been edited 2 times. Last edited by SunnyEvan, Apr 12, 2025, 2:45 PM

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sqing

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#19 h Apr 12, 2025, 3:02 PM • 1 Y

Y by cubres

Nice.Thank SunnyEvan.

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