ineq3

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manlio

3253 posts

manlio

#1 h Jun 6, 2004, 6:04 PM • 8 Y

Y by anhanh123, 277546, bel.jad5, Adventure10, Mango247, and 3 other users

For $a,b,c$ positive real numbers prove that

$a^2+b^2+c^2 \geq \sqrt {3(a^3b+b^3c+c^3a)}$

This post has been edited 1 time. Last edited by manlio, Jun 8, 2004, 10:15 AM

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harazi

5526 posts

harazi

#2 h Jun 6, 2004, 6:12 PM • 2 Y

Y by Adventure10, Mango247

This is a hard inequality. And also note that from this one we can find another tough inequality, posted by me long time ago: if $a^2+b^2+c^2=3$ then $a^3(b+c)+b^3(c+a)+c^3(a+b)\leq 6$ .

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Sasha

129 posts

Sasha

#3 h Jun 6, 2004, 7:25 PM • 2 Y

Y by Adventure10, Mango247

We may consider that $a,b,c\leq1$ . Consider the functions $g(a,b,c)=f(a)=(\sum a^2)^2-3\sum a^3b$ . Then $g(a,b,c)$ has a minimum.
$f'(a)=4a^3-9a^2b+4a(b^2+c^2)-4c^3$ and $f''(a)=12(a^2-\frac32ba+\frac13(b^2+c^2))\geq0$
So $f(a)$ is minimal when $f'(a)=0$ . Analogously for $b$ and $c$ , $g(a,b,c)$ is minimal when
$4a^3-9a^2b+4a(b^2+c^2)-3c^3=4b^3-9b^2c+4b(c^2+a^2)-3a^3=4c^3-9c^2a+4c(a^2+b^2)-3b^3=0$
Summing we get $\sum (a^3+4ab^2-5a^2b)=0$ which holds iff $a=b=c$ .

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harazi

5526 posts

harazi

#4 h Jun 6, 2004, 7:30 PM • 2 Y

Y by Adventure10, Mango247

I don't think the solution is good. How do you know that the minimum when considered as a function in a is attained at the same point as the minimum when considered as a function in b or c. Am I missing something.

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Sasha

129 posts

Sasha

#5 h Jun 6, 2004, 7:45 PM • 2 Y

Y by Adventure10, Mango247

Let g(a,b,c) have minimum in (a0,b0,c0). Assume that f(a) doesn't have minimum in a0 then we can a1 such as f(a1)>f(a0). Then g(a1,b0,c0)>g(a0,b0,c0). Contradiction. So a0 is the minimum of f(a). Consider h(b)=s(c)=g(a,b,c). In the same way we get that h(b) has a minimum in b0 and s(c) has a minimum in c0. Then f'(a0)=h'(b0)=s'(c0)=0.
Is it good now?

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harazi

5526 posts

harazi

#6 h Jun 6, 2004, 7:56 PM • 2 Y

Y by Adventure10, Mango247

I see, you use partial derivatives. Well, in this case it works with the condition to explain very well the existence of the minimum and so on. Anyway, do not take into account what I say now, I'm very tired and I'm surely idiot!

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harazi

5526 posts

harazi

#7 h Jun 6, 2004, 8:04 PM • 2 Y

Y by Adventure10, Mango247

I don't know this method of partial derivatives, but I knew that for proving that it is a minimum it doesn't suffice just to compute the first partial derivatives and then to solve the system. I knew you have to compute the second partial derivatives and to show that some determinants are positive or negative.

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harazi

5526 posts

harazi

#8 h Jun 6, 2004, 8:41 PM • 2 Y

Y by Adventure10, Mango247

Can anyone help me and tell if this method is corect? It sem\ems to easy to be true!

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manlio

3253 posts

manlio

#9 h Jun 6, 2004, 8:49 PM • 2 Y

Y by Adventure10, Mango247

I think that this method isn't correct because to find the min and max of a function of more variables you must use Sylvester theorem which involves the calculus of three determinants composed with partial derivatives.

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Vasc

2861 posts

Vasc

#10 h Jun 8, 2004, 9:15 AM • 8 Y

Y by Supercali, Vimath, Adventure10, Mango247, and 4 other users

Sasha is only partially right.
If $f_1(a)=g(a,b,c)$ is minimal at $a=1$ , $f_2(b)=g(a,b,c)$ is minimal at $b=1$ and $f_3(c)=g(a,b,c)$ is minimal at $c=1$ , then $g(a,b,c)$ is minimal at $(1,1,1)$ .
Indeed, we have $g(a,b,c\geq g(1,b,c) \geq g(1,1,c) \geq g(1,1,1)$ .
But the obtained inequality $g(a,b,c)\geq g(1,1,1)$ is trivial.
In our case, $f_1(a)=(\sum a^2)^2-3\sum a^3b$ is not minimal at $a=1$ for all $b$ and $c$ . Moreover, the conditions $f_1'(1,1,1)=f_2'(1,1,1)=f_3'(1,1,1)=0$ and $(f_1)''(1,1,1)>0$ , $(f_2)''(1,1,1)>0$ , $(f_3)''(1,1,1)>0$
dont imply $g(a,b,c)$ is (local)minimal at (1,1,1). Consequently, the Sasha solution is not correct.
I have to notice that the inequality proposed by Manlio with unknown source was published in G.M 7-8/1992 (problem 22695, pg 287, author Vasc, sorry) in the form

$a^2(a-b)(a-2b)+b^2(b-c)(b-2c)+c^2(c-a)(c-2a)\geq0$ .

This inequality is not trivial, is a very strong inequality. It follows from the identity

$4(a^2+b^2+c^2-ab-bc-ca)[(a^2+b^2+c^2)^2-3(a^3b+b^3c+c^3a)]=$
$=[(a^3+b^3+c^3)-5(a^2b+b^2c+c^2a)+4(ab^2+bc^2+ca^2)]^2+$
$+3[(a^3+b^3+c^3)-(a^2b+b^2c+c^2a)-2(ab^2+bc^2+ca^2)+6abc]^2$ .

I have also a very nice solution for this inequality, but I will not post it, because it is a part of a problem proposed by me to Crux.

This post has been edited 1 time. Last edited by darij grinberg, Dec 30, 2015, 7:55 PM
Reason: faulty latex (\" instead of '')

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harazi

5526 posts

harazi

#11 h Jun 8, 2004, 11:05 AM • 2 Y

Y by Adventure10, Mango247

I have just one simple question: how on earth did you deduce this identity? I mean, I just can't imagine deducing such thing. It is absolutely unbelievable!

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Vasc

2861 posts

Vasc

#12 h Jun 8, 2004, 11:33 AM • 4 Y

Y by Adventure10, Mango247, and 2 other users

This identity follows from the proof.
This one can be an little indication about my proof.

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manlio

3253 posts

manlio

#13 h Jun 8, 2004, 1:43 PM • 3 Y

Y by Adventure10, Mango247, and 1 other user

Vasc, this little clue makes the search of your VERY NICE

proof a fantastic challenge for all very clever solvers of inequalities of the forum.

(unfortunately I am not among them, but I will try it anyway)

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galois

400 posts

galois

#14 h Jun 9, 2004, 10:11 AM • 2 Y

Y by Adventure10, Mango247

again this problem depends on the sign of S=(a-b)(b-c)(c-a).if S is greater than 0 ,then the inequality is proved by what i've done but if S is less than zero then taking a= max{a,b,c} and c<b is th true
the substitution b=b,a=b+ :lambda,c=b- :gamma: kills me with computations
any1 has cum up with anything different ,not vasc's identity,it's a bolt from the blue

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harazi

5526 posts

harazi

#15 h Jun 9, 2004, 10:21 AM • 2 Y

Y by Adventure10, Mango247

My last dream (well, maybe not exactly the last, but close!) is to see a beautiful 3 lines solution without identities. But somthing tells me that such solution can't exist. The inequality is far too strong. Keep searching. After all, for that problem in Crux I spent more than 2 months.

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mudok

3377 posts

mudok

#96 h Oct 14, 2017, 8:15 PM • 3 Y

Y by adityaguharoy, Adventure10, Mango247

Leonard and Kittyw, you are very special mathlinkers, please don't hurt each other!

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janitor

287 posts

janitor

#97 h Jan 29, 2018, 4:58 PM • 1 Y

Y by Adventure10

Original solution belongs to kittyw
$\begin{align*} \ (a^2+b^2+c^2)^2\geq 3(a^3b+b^3c+c^3a)\\ \text{or}\\ \ 2(a^2+b^2+c^2)^2\geq 3(a^3(b+c)+b^3(c+a)+c^3(a+b)-(a+b+c)(a-b)(b-c)(c-a))\\ \text{or}\\ \sum (b^4+c^4-3bc(b^2+c^2)+4b^2c^2)=\sum (b^2-bc+c^2)(b-c)^3\geq 3(-(a+b+c)(a-b)(b-c)(c-a))\\ \text{Let's 3=a+b+c }\ 3v^2=ab+bc+ca \text{ }\ w^3=abc\\ \text{Then inequality equivalent to}\\ \ 18u^4-33u^2v^2+14v^4+uw^3\geq -u(a-b)(b-c)(c-a)\\ \text{Squaring}\\ \ u^2w^6+2(18u^4-33u^2v^2+14v^4)uw^3+324u^8-1188u^6v^2+1593u^4v^4-924u^2v^6+196v^8\geq 27u^2(3u^2v^4-4v^6-4u^3w^3+6uv^2w^3-w^6)\\ \text{or}\\ \ 7u^2w^6+(36u^4-57u^2v^2+7v^4)uw^3+81u^8-297u^6v^2+378u^4v^4-204u^2v^6+49v^8\\ \text{Then enough to prove that }\ \Delta \leq0 \text{ , which is equivalent to }\\ \ 27u^2(36u^8-156u^6v^2+253u^4v^4-182u^2v^6+49v^8)=27u^2(6u^2-7v^2)^2(u^2-v^2)^2\geq0 \end{align*}$

This post has been edited 3 times. Last edited by janitor, Jan 29, 2018, 10:56 PM

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adityaguharoy

4655 posts

adityaguharoy

#98 h Mar 1, 2018, 5:29 PM • 1 Y

Y by Adventure10

There is a related version of it posted under some "I Get it" collection here in this forum on AoPS by Grotex. But can't find it now due to poor searching.

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TaHongQuang

368 posts

TaHongQuang

#99 h Sep 22, 2019, 4:17 AM • 1 Y

Y by Adventure10

THQ202 (Inspired by VASC and Sladjian Stankovik)
Let $a, b, c$ be real numbers and $s = \sqrt {\frac{{{a^2} + {b^2} + {c^2} - ab - bc - ca}}{7}}$ . Prove

$\left( {a}^{2}+{b}^{2}+{c}^{2} \right) ^{2} \ge 3\left( {a}^{3}b+{b}^{3}c+{c}^{3}a \right) +{\frac {7\,{s}^{2} \left( a+b+c-7\,s \right) ^{2}}{9}}\$

This post has been edited 1 time. Last edited by TaHongQuang, Sep 22, 2019, 5:13 AM
Reason: Typo

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arqady

30211 posts

arqady

#100 h Sep 22, 2019, 6:48 AM • 2 Y

Y by TaHongQuang, Adventure10

TaHongQuang wrote:

THQ202
Let $a, b, c$ be real numbers and $s = \sqrt {\frac{{{a^2} + {b^2} + {c^2} - ab - bc - ca}}{7}}$ . Prove

$\left( {a}^{2}+{b}^{2}+{c}^{2} \right) ^{2} \ge 3\left( {a}^{3}b+{b}^{3}c+{c}^{3}a \right) +{\frac {7\,{s}^{2} \left( a+b+c-7\,s \right) ^{2}}{9}}\$

The $uvw$ 's technique should kill it.

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mihaig

7343 posts

mihaig

#101 h Oct 12, 2021, 10:05 PM

Y by

TaHongQuang wrote:

THQ202 (Inspired by VASC and Sladjian Stankovik)
Let $a, b, c$ be real numbers and $s = \sqrt {\frac{{{a^2} + {b^2} + {c^2} - ab - bc - ca}}{7}}$ . Prove

$\left( {a}^{2}+{b}^{2}+{c}^{2} \right) ^{2} \ge 3\left( {a}^{3}b+{b}^{3}c+{c}^{3}a \right) +{\frac {7\,{s}^{2} \left( a+b+c-7\,s \right) ^{2}}{9}}\$

After a few moves (not easy, but not difficult either) we get:
$\min\left[\left( {a}^{2}+{b}^{2}+{c}^{2} \right) ^{2}-3\left( {a}^{3}b+{b}^{3}c+{c}^{3}a \right)\right]=\frac {7\,{s}^{2} \left( a+b+c-7\,s \right) ^{2}}{9}.$ (of course, by fixing first $a+b+c\neq0$ and $ab+bc+ca$ ).
Note that if $a+b+c=0,$ then LHS=RHS. Interesting identity.

This post has been edited 2 times. Last edited by mihaig, Oct 12, 2021, 10:28 PM

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mihaig

7343 posts

mihaig

#102 h Oct 13, 2021, 4:58 AM

Y by

arqady wrote:

TaHongQuang wrote:

THQ202
Let $a, b, c$ be real numbers and $s = \sqrt {\frac{{{a^2} + {b^2} + {c^2} - ab - bc - ca}}{7}}$ . Prove

$\left( {a}^{2}+{b}^{2}+{c}^{2} \right) ^{2} \ge 3\left( {a}^{3}b+{b}^{3}c+{c}^{3}a \right) +{\frac {7\,{s}^{2} \left( a+b+c-7\,s \right) ^{2}}{9}}\$

The $uvw$ 's technique should kill it.

I wrote in October 2017 an article, where I show through different methods that
$\min\left[\left( {a}^{2}+{b}^{2}+{c}^{2} \right) ^{2}-3\left( {a}^{3}b+{b}^{3}c+{c}^{3}a \right)\right]=\frac {7\,{s}^{2} \left( a+b+c-7\,s \right) ^{2}}{9},$ for $a+b+c\neq0$ and $ab+bc+ca,$ both fixed. It will appear this week in GMB (Gazeta Matematica).

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teomihai

2957 posts

teomihai

#103 h Oct 13, 2021, 7:06 AM

Y by

arqady wrote:

manlio wrote:

For $a,b,c$ positive real numbers prove that

$a^2+b^2+c^2 \geq \sqrt {3(a^3b+b^3c+c^3a)}$

Let $a=x+2\cos\frac{\pi}{7}y$ , $b=y+2\cos\frac{\pi}{7}z$ and $c=z+2\cos\frac{\pi}{7}x$ .
Hence, $\sum_{cyc}a^3b=\sum_{cyc}\left(x+2\cos\frac{\pi}{7}y\right)^3\left(y+2\cos\frac{\pi}{7}z\right)=$
$=\sum_{cyc}\left(x^3+6\cos\frac{\pi}{7}x^2y+12xy^2\cos^2\frac{\pi}{7}+8\cos^3\frac{\pi}{7}y^3\right)\left(y+2\cos\frac{\pi}{7}z\right)=$
$=\sum_{cyc}\left(Ax^4+Bx^2y^2+Cx^2yz+\left(1+16\cos^4\frac{\pi}{7}\right)x^3y+\left(2\cos\frac{\pi}{7}+12\cos^2\frac{\pi}{7}\right)x^3z\right)$ ,
where $A$ , $B$ and $C$ are constants.
But $1+16\cos^4\frac{\pi}{7}=2\cos\frac{\pi}{7}+12\cos^2\frac{\pi}{7}$ .
Thus, we need to prove a fourth degree symmetric inequality.
Let $x+y+z=3u$ , $xy+xz+yz=3v^2$ , where $v^2$ can be negative and $xyz=w^3$ .
Id est, we need to prove that $f(w^3)\geq0$ , where $f$ is a linear function.
Hence, $f$ gets a minimal value for an extremal value of $w^3$ , which happens, when two variables are equal.
Since we have a fourth degree inequality, we can assume $y=z=1$ .
Thus, it remains to prove that $\left(\left(x+2\cos\frac{\pi}{7}\right)^2+\left(1+2\cos\frac{\pi}{7}\right)^2+\left(1+2\cos\frac{\pi}{7}x\right)^2\right)^2\geq$
$\geq3\left(\left(x+2\cos\frac{\pi}{7}\right)^3\left(1+2\cos\frac{\pi}{7}\right)+\left(1+2\cos\frac{\pi}{7}\right)^3\left(1+2\cos\frac{\pi}{7}x\right)+\left(1+2\cos\frac{\pi}{7}x\right)^3\left(x+2\cos\frac{\pi}{7}\right)\right)$ ,
which is $(x-1)^2\left(x-8\cos^2\frac{\pi}{7}-2\cos\frac{\pi}{7}+1\right)^2\geq0$ . Done!

beautiful!

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mihaig

7343 posts

mihaig

#104 h Oct 13, 2021, 8:07 AM

Y by

It's wonderful. But you have also access to GMB and you'll soon read my article

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teomihai

2957 posts

teomihai

#105 h Oct 13, 2021, 8:14 AM

Y by

mihaig wrote:

It's wonderful. But you have also access to GMB and you'll soon read my article

Thanks, I look forward to it!

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mihaig

7343 posts

mihaig

#106 h Oct 26, 2021, 9:14 AM

Y by

See my article on this fascinant result of the professor in the below link. By courtesy of the Romanian Mathematical Society (the article is in the official site)
https://ssmr.ro/gazeta/gmb/2021/9/articol.pdf

Attachments:

articol.pdf (203kb)

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teomihai

2957 posts

teomihai

#107 h Oct 26, 2021, 10:49 AM

Y by

indeed splendid article.

This post has been edited 1 time. Last edited by teomihai, Oct 26, 2021, 6:23 PM

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edemafa

212 posts

edemafa

#108 h Oct 26, 2021, 4:50 PM

Y by

manlio wrote:

For $a,b,c$ positive real numbers prove that

$a^2+b^2+c^2 \geq \sqrt {3(a^3b+b^3c+c^3a)}$

We gef

$a^2+b^2+c^2 \geq \sqrt {3(a^3b+b^3c+c^3a)}$ $\implies a^2(a^2+2b^2-3ab) +b^2(b^2+2c^2-3bc) + c^2 (c^2 +2a^2 - 3ac) \geq 0$
WLOG we can suppose $a= min(a,b,c)$ so:

$\implies a^2(a^2+2b^2-3ab) +b^2(b^2+2c^2-3bc) + c^2 (c^2 +2a^2 - 3ac) \geq 3a^2(a^2+b^2+c^2 -ab-ac-bc) \geq 0$
with equality if $a=b=c$

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mihaig

7343 posts

mihaig

#109 h Oct 26, 2021, 6:17 PM • 1 Y

Y by teomihai

teomihai wrote:

indeed splendid articol.

Thank you, dear professor. Not only solves the problem, but it refine it. I will mention again that the article was sent in 2017. I'm sure that meanwhile new results appeared

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Rhapsodies_pro

481 posts

Rhapsodies_pro

#110 h Aug 12, 2022, 4:35 AM • 2 Y

Y by teomihai, Mango247

Some other identities.
Something stranger: For $a,b,c>0$ , show that $7\sum_{\rm cyc}{a^4bc}+\sum_{\rm cyc}{a^3b^3}-29\sum_{\rm cyc}{a^3b^2c}+12\sum_{\rm cyc}{a^3bc^2}+9\sum_{\rm cyc}{a^2b^2c^2}\ge0$ and $\sum_{\rm cyc}{a^5}-\sum_{\rm cyc}{a^4b}-2\sum_{\rm cyc}{a^4c}+3\sum_{\rm cyc}{a^3b^2}-\sum_{\rm cyc}{a^3c^2}+2\sum_{\rm cyc}{a^3bc}-2\sum_{\rm cyc}{a^2b^2c}\ge0\text.$

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