stronger than old result

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mudok

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mudok

#1 h Apr 11, 2016, 11:36 AM • 5 Y

Y by luofangxiang, Misha57, bitrak, Adventure10, Mango247

$a,b,c\ge 0, \ \ a^2+b^2+c^2+abc=4$ . Prove that $a+b+c\ge 2+\sqrt{abc(4-a-b-c)}$
old result

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bitrak

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#2 h Apr 12, 2016, 9:13 PM • 3 Y

Y by Stuart111, Adventure10, Mango247

My solution for this beautiful problem.

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bitrak

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#3 h Apr 12, 2016, 9:16 PM • 1 Y

Y by Adventure10

For younger.
https://www.facebook.com/photo.php?fbid=10154076651786552&set=g.1486244404996949&type=1&theater

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mudok

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#5 h Apr 13, 2016, 4:36 AM • 2 Y

Y by Adventure10, Mango247

See the proof for $a,b,c\ge 0, a^2+b^2+c^2+abc=4 \Longrightarrow a+b+c\ge 2+\sqrt{abc}$
http://artofproblemsolving.com/community/c6h208078p1145009

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arqady

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arqady

#6 h Apr 13, 2016, 5:40 AM • 3 Y

Y by mudok, Adventure10, Mango247

Dear bitrak. In your reasoning we can't say that $F$ is a function with one variable $r$ because $r$ depends on $p$ and $q$ .

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luofangxiang

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luofangxiang

#7 h Apr 13, 2016, 5:50 AM • 4 Y

Y by Stuart111, sabkx, Adventure10, Mango247

//cdn.artofproblemsolving.com/images/8/b/1/8b1be393b296e4482f00265b4e0cd91da4e08bbd.jpg

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mudok

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#8 h Apr 13, 2016, 6:13 AM • 3 Y

Y by sabkx, Adventure10, Mango247

Brilliant proof , luofangxiang !

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bitrak

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#9 h Apr 13, 2016, 7:55 AM • 1 Y

Y by Adventure10

Dear Argady, what you know about ABC Theorem ? Would be nice if you send us links of solved examples with ABC Theorem.

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#10 h Apr 13, 2016, 8:18 AM • 2 Y

Y by Adventure10, Mango247

For younger and older.
One more example of using ABC Theorem.

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bitrak

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#12 h Apr 13, 2016, 9:56 AM • 2 Y

Y by Adventure10, Mango247

And one more inequality can be solved with ABC Theorem.
Now to see that in constraint we have r = f(p)

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Monreal

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Monreal

#13 h Apr 13, 2016, 10:13 AM • 2 Y

Y by Adventure10, Mango247

bitrak wrote:

Dear Argady, what you know about ABC Theorem ? Would be nice if you send us links of solved examples with ABC Theorem.

Dear bitrak, Arqady right, this problem can be solved by ABC theorem because $r$ depend on $p$ and $q$ , you can read the $uvw$ theorem ( or ABC theorem, I learn it from Nguyen Anh Cuong) and I sure that your solution has wrong already.

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bitrak

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#14 h Apr 13, 2016, 10:27 AM • 1 Y

Y by Adventure10

Monreal, you know Vietnamese language ^^.
UVW and ABC are not same technique (methods) such that maybe you don't learn good.
I expect comments from others mathematicians.
Thanks for your comment Monreal.

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bitrak

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#15 h Apr 13, 2016, 10:45 AM • 1 Y

Y by Adventure10

luofangxiang
It is true that:
c ≥ 1 ( must be proved: Because c=max{a,b,c} => c²+c²+c²+c³ ≥ 4 <=> (c-1)(c+2)² ≥0 <=> c ≥1 ) and
c ≥ (a+b)/2 (proof: from c ≥a and c ≥ b => 2c ≥ (a+b) => c ≥ (a+b)/2 )
How from here you conclude that
c ≥ 1≥ (a+b)/2
?
Thanks for your explanation.

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luofangxiang

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luofangxiang

#16 h Apr 13, 2016, 10:49 AM • 2 Y

Y by Adventure10, Mango247

a+b+c<=3

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bitrak

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#17 h Apr 13, 2016, 10:51 AM • 2 Y

Y by Adventure10, Mango247

luofangxiang
But and a+b+c ≤ 3 must be proved, no?

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arqady

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#18 h Apr 13, 2016, 10:51 AM • 2 Y

Y by Adventure10, Mango247

bitrak wrote:

UVW and ABC are not same technique (methods)

I think, they are the same.
We can not use $uvw$ (at least, I don't see how we can use) for the proof of the mudok's inequality.

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bitrak

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#19 h Apr 13, 2016, 11:02 AM • 2 Y

Y by Adventure10, Mango247

According it that the proofs of UVW and ABC are different and not similar
I think that these techniques are not same.

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mudok

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mudok

#20 h Apr 13, 2016, 11:07 AM • 2 Y

Y by Adventure10, Mango247

bitrak wrote:

a+b+c ≤ 3 must be proved, no?

It is well-known result. For the proof, for example, see here:

http://artofproblemsolving.com/community/c6h506413p2844837

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mudok

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#21 h Apr 13, 2016, 11:29 AM • 2 Y

Y by Adventure10, Mango247

bitrak wrote:

According it that the proofs of UVW and ABC are different and not similar
I think that these techniques are not same.

To figure out it we should see ABC theorem and its proof.

About $uvw$ you can see here:
http://artofproblemsolving.com/community/c6h278791p1507763

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bitrak

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#22 h Apr 13, 2016, 11:29 AM • 2 Y

Y by Adventure10, Mango247

Nice mudok and thank you for helping in explanation the solution of luofangxiang.
Your proof for a+b+c ≤ 3 is very nice. I have this proof from year 2011 on vietnamese language.
That's first line of his proof and the most important fact.
" It is well-known result ", but in solution of luofangxiang must be noted and prove.
^^

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Reason: I forgot a picture.

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mudok

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#25 h Apr 13, 2016, 11:58 AM • 2 Y

Y by Adventure10, Mango247

bitrak wrote:

" It is well-known result ", but in solution of luofangxiang must be noted and prove.

ABC theorem is not "well-known". In your solution it must be noted and proved.

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bitrak

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#26 h Apr 13, 2016, 2:58 PM • 2 Y

Y by Adventure10, Mango247

You are right mudok.
On AoPS ABC theorem is not "well-known result ".
Same in Europe.
I think that ABC Theorem is before UVW and is proved by Vietnamese mathematicians,
because very often I see proof written on vietnamese language (almost never on englesh).
But here on AoPS have so many good mathematicians from Asia and they can help us with translating and elaborating of ABC theorem.
In each case, by my thinking ABC is very useful tool for solving inequalities as UVW method.

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mudok

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#27 h Apr 13, 2016, 3:34 PM • 1 Y

Y by Adventure10

For example, why not you translate it to english?

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bitrak

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#28 h Apr 13, 2016, 3:45 PM • 2 Y

Y by Adventure10, Mango247

Ha

I don't know vietnamese language.

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mudok

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#29 h Apr 13, 2016, 3:51 PM • 1 Y

Y by Adventure10

bitrak wrote:

Ha

I don't know vietnamese language.

how did you know that $ABC$ and $uvw$ are different?

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bitrak

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#30 h Apr 13, 2016, 4:04 PM • 1 Y

Y by Adventure10

haha ;

that's mathematics, numbers and letters ^^
I will send one document later on vietnamese language sure, with proof.

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#32 h Apr 13, 2016, 8:22 PM • 2 Y

Y by Adventure10, Mango247

mudok wrote:

bitrak wrote:

a+b+c ≤ 3 must be proved, no?

It is well-known result. For the proof, for example, see here:

http://artofproblemsolving.com/community/c6h506413p2844837

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urun

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urun

#33 h Apr 13, 2016, 9:57 PM • 3 Y

Y by kilua, Adventure10, Mango247

$uvw$ and $ABC$ techniques are the same. The original proof, written in Vietnamese, is essentially the same as one written in English. So you should not argument about it, Bitrak .

Argady is correct when arguing that you cannot fix $a+b+c$ and $ab+bc+ca$ constant at the same time and treat $abc$ as a single variable. The core ideal of $ABC$ theorem is to expand $P=(a-b)^2(b-c)^2(c-a)^2$ to find the bound of $abc$ . Then it is arguing that any symmetric polynomial that can be written as monotonic function of $abc$ will achieve extrema when 2 variables are equal. Anhcuong simplied the criterion by considering any symmetric polynomial of degree $<= 5$ because obviously they are all linear function of $abc$ .

It is obviously that you did not understand the proof of $ABC$ or $uvw$ theorem .

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#34 h Apr 13, 2016, 10:44 PM • 2 Y

Y by xlzq, Adventure10

I think Urun that you don't understand ABC theorem. In my solved example above you can see that p and q are not fixed and ABC technique is explained complete. So, would be nice if we can see the proof of ABC on english and Vietnamese language.
Your comment is very suspicious without proof.
Thank you for a comment.

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mudok

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#35 h Apr 14, 2016, 3:14 AM • 2 Y

Y by Adventure10, Mango247

bitrak wrote:

Your comment is very suspicious without proof.

Sorry, Bitrak, but there is only one person who speaks very suspicious without proof.

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bitrak

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#36 h Apr 14, 2016, 7:54 AM • 2 Y

Y by mudok, Adventure10

Because problem of ABC and UVW is very interesting and behind your very nice inequlaity, I will open new topics where we can
continue with comments. I expect for your inequality here will have more nice solutions.
If you are the author of the inequality, then bravo for you, inequality is really beautiful.
http://artofproblemsolving.com/community/c6h1227738_abc_vs_uvw_theorem

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Misha57

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#37 h Apr 14, 2016, 8:53 AM • 2 Y

Y by Adventure10, Mango247

.

mudok wrote:

$a,b,c\ge 0, \ \ a^2+b^2+c^2+abc=4$ . Prove that $a+b+c\ge 2+\sqrt{abc(4-a-b-c)}$
old result

Lets fix $w^3$ .We have condition $9u^2-6v^2+w^3=4 <==> v^2=1/6(9u^2+w^3-4)$ .We see that inequality is equivalent to $f(u)\ge 0$ where $f(x)$ is $3x-2-L*\sqrt{4-3x}$ where $L=\sqrt{abc}-constant.$ We have $f'=\frac{3L}{2\sqrt{4-3x}}+3>0$ so $f$ is monotonic.So we only need to check for minimal possible value of u.Existence of a,b,c for u is equivalent to $-(w^3-3uv^2+2u^3)^2+4(u^2-v^2)\ge 0$ <==> $-(const_1+const_2v+const_3v^2-5/2u^3)^2+4(-1/2v^2+const_4v+const_5)^3\ge 0$ <==> $P(u)=Au^6+Bu^5+...+G\ge0$ where $A=-25/4-1/2<0$ ,so we see that for big enough $x$ $P(x)<0$ and for some $x$ $P(x)\ge0$ so it has smallest positive real root $k$ .If $p(x) <0$ when $x\in [0,k)$ then smallest value u can attain is $k$ otherwise its $0$ .In both cases two of $a,b,c$ are equal,so we only need to check inequality when $a=b$ , and this case is obvious.
Note that we allow $v^2$ to be negative(but not $u,w^3$ )(because in case $a=b$ inequality is still true(see #2 post)
So inequality is true for all reals $a,b,c$ such that $a+b+c,abc\ge0,a^2+b^2+c^2+abc=4$ .We also have one more equality case for this: $(-2,0,0)$

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mudok

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#38 h Apr 14, 2016, 9:48 AM • 2 Y

Y by Adventure10, Mango247

Misha57 wrote:

$f(x)$ is $3x-2-L*\sqrt{4-3x}$ where $L=\sqrt{abc}-constant.$

By the condition, $L^2=abc$ depends on $x$ , so, is $L$ constant?

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Misha57

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#39 h Apr 14, 2016, 9:50 AM • 2 Y

Y by Adventure10, Mango247

We fix any $w^3$ for which there exist $u,v^2$ such that $9u^2-6v^2+w^3=4.$ Then we are moving u (with fixed $w^3$ and condition $v^2=1/6(9u^2+w^3-4)$ ).So by each value of $u$ that we chose we know exactly $v^2$ (and $w^3$ since it is fixed).

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mihaig

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#40 h Aug 13, 2022, 10:37 AM • 3 Y

Y by Mango247, Mango247, Mango247

mudok wrote:

$a,b,c\ge 0, \ \ a^2+b^2+c^2+abc=4$ . Prove that $a+b+c\ge 2+\sqrt{abc(4-a-b-c)}$
old result

Sketch. If $abc=0$ then we are done. If $abc>0,$ then for fixed $a^2+b^2+c^2$ and $abc,$
$a+b+c$ is minimum for $a=b\leq c.$ Hence replacing $a=b=x\in(0,1]$ and $c=2-x^2,$ we get
$x^4(x-1)^2\geq0,$ which is true.

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mihaig

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#41 h Aug 14, 2022, 4:35 AM

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