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A cyclic one

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inequalities proposed

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Source: Proposed

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#1 h Sep 15, 2018, 3:48 PM • 2 Y

Y by Adventure10, Mango247

Let $a,b,c>0$ such that $a+b+c=3$ . Prove that:
$\frac{a}{b}+ \frac{b}{c}+ \frac{c}{a} \ge \frac{15}{{3abc+2}}$

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arqady

#2 h Sep 15, 2018, 6:50 PM • 2 Y

Y by Adventure10, Mango247

KaiRain wrote:

Let $a,b,c>0$ such that $a+b+c=3$ . Prove that:
$\frac{a}{b}+ \frac{b}{c}+ \frac{c}{a} \ge \frac{15}{{3abc+2}}$

BW kills it.
About BW see here:
https://artofproblemsolving.com/community/c6h522084

This post has been edited 1 time. Last edited by arqady, Sep 15, 2018, 7:06 PM

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crezk

#3 h Sep 15, 2018, 6:55 PM • 2 Y

Y by Adventure10, Mango247

arqady wrote:

KaiRain wrote:

Let $a,b,c>0$ such that $a+b+c=3$ . Prove that:
$\frac{a}{b}+ \frac{b}{c}+ \frac{c}{a} \ge \frac{15}{{3abc+2}}$

BW kills it.

What is The BW?

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#4 h Sep 15, 2018, 7:08 PM • 1 Y

Y by Adventure10

Read the post above you.

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#5 h Sep 16, 2018, 3:05 AM • 2 Y

Y by Adventure10, Mango247

arqady wrote:

KaiRain wrote:

Let $a,b,c>0$ such that $a+b+c=3$ . Prove that:
$\frac{a}{b}+ \frac{b}{c}+ \frac{c}{a} \ge \frac{15}{{3abc+2}}$

BW kills it.
About BW see here:
https://artofproblemsolving.com/community/c6h522084

Thank you.
How about the following inequality ?

Let $a,b,c>0$ such that $a+b+c=3$ . Prove that:
$\frac{a}{b}+ \frac{b}{c}+ \frac{c}{a} \ge \frac{30}{{1+9\sqrt{abc}}}$

This post has been edited 5 times. Last edited by KaiRain, Sep 16, 2018, 3:36 AM

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#6 h Sep 16, 2018, 3:35 AM • 2 Y

Y by Adventure10, Mango247

Do you have another solution for those inequalities, without lots of calculation ?

This post has been edited 1 time. Last edited by KaiRain, Sep 16, 2018, 3:38 AM

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arqady

#7 h Sep 16, 2018, 5:03 AM • 2 Y

Y by KaiRain, Adventure10

For the proof of both inequalities we can use also the $uvw$ 's technique. For the first turns out even something nice.

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30171 posts

arqady

#8 h Sep 16, 2018, 6:31 AM • 2 Y

Y by KaiRain, Adventure10

KaiRain wrote:

Let $a,b,c>0$ such that $a+b+c=3$ . Prove that:
$\frac{a}{b}+ \frac{b}{c}+ \frac{c}{a} \ge \frac{15}{{3abc+2}}$

I think the following inequality is very nice.
Let $a$ , $b$ and $c$ be positive numbers such that $a+b+c=3$ . Prove that:
$\frac{a}{b}+ \frac{b}{c}+ \frac{c}{a} \ge \frac{12}{3abc+1}.$

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#9 h Sep 16, 2018, 10:14 AM • 2 Y

Y by Adventure10, Mango247

arqady wrote:

I think the following inequality is very nice.
Let $a$ , $b$ and $c$ be positive numbers such that $a+b+c=3$ . Prove that:
$\frac{a}{b}+ \frac{b}{c}+ \frac{c}{a} \ge \frac{12}{3abc+1}.$

yep!
How can you prove it

This post has been edited 1 time. Last edited by KaiRain, Sep 16, 2018, 10:16 AM

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#11 h Sep 16, 2018, 1:40 PM • 1 Y

Y by Adventure10

I suggest we should find a "human" solution instead of "buffalo" ...

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#12 h Sep 16, 2018, 2:10 PM • 2 Y

Y by Adventure10, Mango247

arqady wrote:

KaiRain wrote:

Let $a,b,c>0$ such that $a+b+c=3$ . Prove that:
$\frac{a}{b}+ \frac{b}{c}+ \frac{c}{a} \ge \frac{15}{{3abc+2}}$

I think the following inequality is very nice.
Let $a$ , $b$ and $c$ be positive numbers such that $a+b+c=3$ . Prove that:
$\frac{a}{b}+ \frac{b}{c}+ \frac{c}{a} \ge \frac{12}{3abc+1}.$

We can write as $a^2b+b^2c+c^2a \ge \frac{12abc}{3abc+1}$ , or $3+\frac{1}{abc} \ge \frac{12}{a^2b+b^2c+c^2a}.$ , but it is https://artofproblemsolving.com/community/q1h618791p3690753

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arqady

#13 h Sep 16, 2018, 8:10 PM • 2 Y

Y by Adventure10, Mango247

I found very nice solution by AM-GM (even twice!).
The equality occurs for $a=b=c=1$ and again for $a$ , $b$ and $c$ are roots of the equation:
$x^3-3x^2+2x-\frac{1}{3}=0.$

This post has been edited 1 time. Last edited by arqady, Sep 16, 2018, 8:13 PM

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#14 h Sep 17, 2018, 12:19 PM • 1 Y

Y by Adventure10

@leon.tymen, arqady hasn't replied yet ...

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#15 h Sep 17, 2018, 3:03 PM • 2 Y

Y by Adventure10, Mango247

arqady wrote:

I found very nice solution by AM-GM (even twice!).
The equality occurs for $a=b=c=1$ and again for $a$ , $b$ and $c$ are roots of the equation:
$x^3-3x^2+2x-\frac{1}{3}=0.$

That's great

Arqady, can you post your solution, or give us a hint? Thank you

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30171 posts

arqady

#16 h Sep 17, 2018, 6:31 PM • 3 Y

Y by NaPrai, Adventure10, Mango247

OK! Since this problem is very old, I'll post my solution.
Let $a$ , $b$ and $c$ be positive numbers such that $a+b+c=3$ . Prove that:
$\frac{a}{b}+ \frac{b}{c}+ \frac{c}{a} \ge \frac{12}{3abc+1}.$ Proof.
Let $a+b+c=3u$ , $ab+ac+bc=3v^2$ , $abc=w^3$ and $u^3=tw^3$ .
Thus, we need to prove that
$\sum_{cyc}a^2c\geq\frac{12u^3w^3}{u^3+3w^3}$ or
$\sum_{cyc}(a^2b+a^2c)-\frac{24u^3w^3}{u^3+3w^3}\geq\sum_{cyc}(a^2b-a^2c)$ or
$9uv^2-3w^3-\frac{24u^3w^3}{u^3+3w^3}\geq(a-b)(a-c)(b-c).$ We'll prove that
$9uv^2-3w^3-\frac{24u^3w^3}{u^3+3w^3}\geq0.$ Indeed, this inequality is equivalent to $f(w^3)\geq0,$ where $f$ is a concave function, which says that it's enough to prove this inequality
for an extreme value of $w^3$ , which happens in the following cases.
1. $w^3\rightarrow0^+$ . Here it's obvious;
2. $b=c=1$ because the last inequality is homogeneous.
Thus, we need to prove that
$(a+2)(2a+1)-3a-\frac{\frac{8(a+2)^3a}{9}}{\frac{(a+2)^3}{27}+3a}\geq0$ or
$(a-1)^2(a^3-3a^2+21a+8)\geq0.$ Id est, it's enough to prove that
$\left(9uv^2-3w^3-\frac{24u^3w^3}{u^3+3w^3}\right)^2\geq\prod_{cyc}(a-b)^2$ or
$3\left(uv^2-\frac{3u^3w^3+w^6}{u^3+3w^3}\right)^2\geq3u^2v^4-4v^6-4u^3w^3+6uv^2w^3-w^6$ or
$4v^6-6uv^2w^3\left(1+\frac{3u^3+w^3}{u^3+3w^3}\right)+\frac{3(3u^3+w^3)^2w^6}{(u^3+2w^3)^2}+w^6+4u^3w^3\geq0$ or
$v^6+\frac{(7u^6+6u^3w^3+3w^6)w^6}{(u^3+3w^3)^2}+u^3w^3\geq\frac{6uv^2w^3(u^3+w^3)}{u^3+3w^3}.$ Now, by AM-GM
$v^6+\frac{(7u^6+6u^3w^3+3w^6)w^6}{(u^3+3w^3)^2}+u^3w^3\geq3v^2\sqrt[3]{\frac{1}{4}\left(\frac{(7u^6+6u^3w^3+3w^6)w^6}{(u^3+3w^3)^2}+u^3w^3\right)^2}.$ Hence, it's enough to prove that
$\left(\frac{7t^2+6t+3}{(t+3)^2}+t\right)^2\geq\frac{32t(t+1)^3}{(t+3)^3}$ or
$(t^3+13t^2+15t+3)^2\geq32t(t+1)^3(t+3)$ or
$(t^2+12t+3)^2\geq32t(t+1)(t+3),$ which is true by AM-GM again:
$(t^2+12t+3)^2=(t^2+4t+3+8t)^2\geq$ $\geq\left(2\sqrt{8t(t^2+4t+3)}\right)^2=32t(t+1)(t+3).$ Done!

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#17 h Sep 18, 2018, 1:39 AM • 2 Y

Y by Adventure10, Mango247

Thank you a lot, arqady !

This post has been edited 2 times. Last edited by KaiRain, Sep 18, 2018, 1:41 AM

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