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Source: vietnam

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71 posts

hitek

#1 h May 14, 2006, 11:02 AM • 2 Y

Y by Adventure10, Mango247

1) Prove that if $abc=1$ , then
$\frac1{a^2-a+1}+\frac1{b^2-b+1}+\frac1{c^2-c+1}\leq3.$
2) Let $x,y,z$ in $[-1;1]$ , and $x+y+z\geq0$ , prove that
$\sqrt{1+x+\tfrac79x^2}+\sqrt{1+y+\tfrac79y^2}+\sqrt{1+z+\tfrac79z^2}\geq3.$

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91 posts

eLeM

#2 h May 14, 2006, 2:26 PM • 1 Y

Y by Adventure10

hitek wrote:

1) Prove that if $abc=1$ , then
$\frac1{a^2-a+1}+\frac1{b^2-b+1}+\frac1{c^2-c+1}\leq3.$
2) Let $x,y,z$ in $[-1;1]$ , and $x+y+z\geq0$ , prove that
$\sqrt{1+x+\tfrac79x^2}+\sqrt{1+y+\tfrac79y^2}+\sqrt{1+z+\tfrac79z^2}\geq3.$

1) Inequality is equal to:
$\frac{a^2-a}{a^2-a+1}+\frac{b^2-b}{b^2-b+1}+\frac{c^2-c}{c^2-c+1}=\sum \frac{a^2-a}{a^2-a+1}\geq0$

Atfer multiplication it becomes:
$\sum \left(a^2b^2+a^2+b+c-ab-bc-ca-\frac{b}{c}\right)\geq0$ and we have to prove:
$\sum \frac{1}{a^2} + \sum a^2 + 2\sum a\geq \sum \frac{b}{c} + 3\sum ab$

$\sum a^2 \geq \sum \frac{b}{c}$ using Rearrangement and
$\sum \frac{1}{a^2} + 2\sum \frac{1}{bc} = \left(\sum \frac{1}{a}\right)^2\geq3\sum\frac{1}{a}$ using AM-GM.

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30161 posts

arqady

#3 h May 15, 2006, 9:21 PM • 2 Y

Y by Adventure10, Mango247

eLeM wrote:

$\sum a^2 \geq \sum \frac{b}{c}$ using Rearrangement

Try $a=b=2,$ $c=\frac{1}{4}.$

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91 posts

eLeM

#4 h May 15, 2006, 9:41 PM • 2 Y

Y by Adventure10, Mango247

arqady wrote:

eLeM wrote:

$\sum a^2 \geq \sum \frac{b}{c}$ using Rearrangement

Try $a=b=2,$ $c=\frac{1}{4}.$

Thanks for your remark. I have written it in a hury. Moreover, it looks incorrect at the first sight, actually

But I think all is correct until that step so try to find good sulution to it

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30161 posts

arqady

#5 h Jun 17, 2006, 7:39 AM • 3 Y

Y by Pirkuliyev Rovsen, Adventure10, Mango247

hitek wrote:

1) Prove that if $abc=1$ , then
$\frac1{a^2-a+1}+\frac1{b^2-b+1}+\frac1{c^2-c+1}\leq3.$

Let $a<0$ and $b<0.$ Then let $a'=-a,$ $b'=-b.$
Hence, $a'b'c=1$ and
$\frac1{a^2-a+1}+\frac1{b^2-b+1}+\frac1{c^2-c+1}=\frac1{a'^2+a'+1}+\frac1{b'^2+b'+1}+\frac1{c^2-c+1}<$
$<\frac1{a'^2-a'+1}+\frac1{b'^2-b'+1}+\frac1{c^2-c+1}.$
Hence, remain to prove that
$\frac1{a^2-a+1}+\frac1{b^2-b+1}+\frac1{c^2-c+1}\leq3$ for all positive $a,$ $b$ and $c,$ such that $abc=1.$
But $\frac1{a^2-a+1}+\frac1{b^2-b+1}+\frac1{c^2-c+1}\leq3\Leftrightarrow$
$\Leftrightarrow\sum_{cyc}(2a^2b^2-2a^2b-2a^2c+a^2-ab+2a)\geq0\Leftrightarrow$
$\Leftrightarrow(a-1)^2(b-1)^2(c-1)^2+\sum_{cyc}(a^2b^2+2-3ab)\geq0.$
Let $ab=z,$ $ac=y$ and $bc=x.$
Hence, remain to prove that $x^2+y^2+z^2+6\geq3(x+y+z),$ where $xyz=1.$
Let $x=\min\{x,y,z\}$ and $f(x,y,z)=x^2+y^2+z^2+6-3(x+y+z).$ Then $yz\geq1.$
Hence, $f(x,y,z)-f(x,\sqrt{yz},\sqrt{yz})=(y-z)^2-3(\sqrt{y}-\sqrt{z})^2=$
$=(\sqrt y-\sqrt z)^2((\sqrt y+\sqrt z)^2-3)\geq0,$ since $yz\geq1.$
Id est, remain to prove that $f(x,y,y)\geq0,$ where $y\geq1.$
But $f(x,y,y)\geq0\Leftrightarrow(y-1)^2(2y^4-2y^3+2y+1)\geq0.$

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41319 posts

sqing

#6 h Dec 7, 2019, 4:23 AM • 2 Y

Y by Adventure10, Mango247

hitek wrote:

Prove that if $abc=1$ , then
$\frac1{a^2-a+1}+\frac1{b^2-b+1}+\frac1{c^2-c+1}\leq3.$

Prove that if $abc=1$ , then $\frac1{a^2+a+1}+\frac1{b^2+b+1}+\frac1{c^2+c+1}\ge1$ $\frac1{a^2+a}+\frac1{b^2+b}+\frac1{c^2+c}\ge \frac{3}{2}$ Old.
Where?

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1027 posts

#7 h Aug 25, 2024, 3:27 AM

Y by

hitek wrote:

1) Prove that if $abc=1$ , then
$\frac1{a^2-a+1}+\frac1{b^2-b+1}+\frac1{c^2-c+1}\leq3.$ 2) Let $x,y,z$ in $[-1;1]$ , and $x+y+z\geq0$ , prove that
$\sqrt{1+x+\tfrac79x^2}+\sqrt{1+y+\tfrac79y^2}+\sqrt{1+z+\tfrac79z^2}\geq3.$

How to prove the second problem?

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30161 posts

arqady

#8 h Aug 25, 2024, 4:07 AM • 1 Y

Y by teomihai

jokehim wrote:

hitek wrote:

2) Let $x,y,z$ in $[-1;1]$ , and $x+y+z\geq0$ , prove that
$\sqrt{1+x+\tfrac79x^2}+\sqrt{1+y+\tfrac79y^2}+\sqrt{1+z+\tfrac79z^2}\geq3.$

How to prove the second problem?

Because $\sum_{cyc}\sqrt{1+x+\tfrac79x^2}-3=\sum_{cyc}\left(\sqrt{1+x+\tfrac79x^2}-1\right)=\sum_{cyc}\frac{x+\frac{7}{9}x^2}{\sqrt{1+x+\tfrac79x^2}+1}=$ $=\sum_{cyc}\left(\frac{x+\frac{7}{9}x^2}{\sqrt{1+x+\tfrac79x^2}+1}-\frac{1}{2}x\right)+\frac{x+y+z}{2}=\sum_{cyc}\frac{x\left(\frac{14}{9}x+1-\sqrt{1+x+\frac{7}{9}x^2}\right)}{2\left(\sqrt{1+x+\tfrac79x^2}+1\right)}+\frac{x+y+z}{2}=$ $=\sum_{cyc}\frac{x^2\left(\frac{14}{9}-\frac{1+\frac{7}{9}x}{\sqrt{1+x+\tfrac79x^2}+1}\right)}{2\left(\sqrt{1+x+\tfrac79x^2}+1\right)}+\frac{x+y+z}{2}\geq0$

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30161 posts

arqady

#9 h Aug 25, 2024, 4:22 AM • 3 Y

Y by MihaiT, ehuseyinyigit, teomihai

Another way:
$\sum_{cyc}\sqrt{1+x+\tfrac79x^2}\geq\sum_{cyc}\sqrt{1+x+\frac{1}{4}x^2}=\frac{6+x+y+z}{2}\geq3.$

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744 posts

MihaiT

#11 h Aug 26, 2024, 5:11 PM

Y by

arqady wrote:

Another way:
$\sum_{cyc}\sqrt{1+x+\tfrac79x^2}\geq\sum_{cyc}\sqrt{1+x+\frac{1}{4}x^2}=\frac{6+x+y+z}{2}\geq3.$

Wow!....

:first:

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3301 posts

#12 h Aug 27, 2024, 1:10 AM

Y by

I think the original inequality should be
$\sqrt{1+x+\tfrac79y^2}+\sqrt{1+y+\tfrac79z^2}+\sqrt{1+z+\tfrac79x^2} \geqslant 3.$

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1027 posts

#13 h Aug 28, 2024, 9:57 AM

Y by

Nguyenhuyen_AG wrote:

I think the original inequality should be
$\sqrt{1+x+\tfrac79y^2}+\sqrt{1+y+\tfrac79z^2}+\sqrt{1+z+\tfrac79x^2} \geqslant 3.$

It is also easy

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52 posts

#14 h Aug 28, 2024, 11:13 AM

Y by

easy how

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30161 posts

arqady

#15 h Aug 29, 2024, 6:42 AM

Y by

Nguyenhuyen_AG wrote:

I think the original inequality should be
$\sqrt{1+x+\tfrac79y^2}+\sqrt{1+y+\tfrac79z^2}+\sqrt{1+z+\tfrac79x^2} \geqslant 3.$

I have a proof for $x+y+z=0$ .

This post has been edited 1 time. Last edited by arqady, Aug 29, 2024, 8:04 AM

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