IMO ShortList 2001, algebra problem 6

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orl

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orl

#1 h Sep 30, 2004, 5:19 PM • 10 Y

Y by Adventure10, megarnie, son7, HWenslawski, lian_the_noob12, Mango247, NicoN9, and 3 other users

Prove that for all positive real numbers $a,b,c$ , $\frac{a}{\sqrt{a^2 + 8bc}} + \frac{b}{\sqrt{b^2 + 8ca}} + \frac{c}{\sqrt{c^2 + 8ab}} \geq 1.$

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Rafal

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Rafal

#2 h Nov 6, 2004, 9:26 PM • 24 Y

Y by cindy7, FlakeLCR, biomathematics, YadisBeles, me9hanics, ValidName, bobjoe123, HolyMath, Illuzion, Adventure10, myh2910, megarnie, son7, Vladimir_Djurica, ehuseyinyigit, Mango247, bhan2025, Sedro, volca_, and 5 other users

this inequality is homogeneous so we can put $a+b+c=1$
and apply Jensen's inequality for $f(x)=\frac{1}{\sqrt{x}}$
so we have :
$\frac{a}{\sqrt{a^2+8bc}}+\frac{b}{\sqrt{b^2+8ac}}+\frac{b}{\sqrt{b^2+8ac}} \geq \frac{1}{\sqrt{a^3+b^3+c^3+24abc}}$
but :
$1=(a+b+c)^3=a^3+b^3+c^3+6abc+3(a^2b+a^2c+b^2a+b^2c+c^2a+c^2b) \geq a^3+b^3+c^3+24abc$

, so indeed $\frac{1}{\sqrt{a^3+b^3+c^3+24abc}} \geq 1$ .

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Karth

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Karth

#3 h Sep 22, 2006, 11:14 PM • 10 Y

Y by Adventure10, megarnie, son7, Mango247, volca_, and 5 other users

Sorry for reviving this post, firstly. I had a question about this inequality. Why does $a+b+c = 1$ ? Why can we do this?

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rem

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rem

#4 h Sep 22, 2006, 11:25 PM • 12 Y

Y by Learner94, SidVicious, me9hanics, Happy2020, Adventure10, megarnie, son7, Vladimir_Djurica, Mango247, volca_, and 2 other users

Because the inequality is homogenous.
Let's say $a+b+c=k$
Taking $a'=a/k, b'=b/k, c'=c/k$ would give the same result but $a'+b'+c'=1$

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Phelpedo

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Phelpedo

#5 h Apr 21, 2007, 11:59 PM • 6 Y

Y by me9hanics, Adventure10, son7, Mango247, and 2 other users

Rafal wrote:

this inequality is homogeneous so we can put $a+b+c=1$
and apply Jensen's inequality for $f(x)=\frac{1}{\sqrt{x}}$
so we have :
$\frac{a}{\sqrt{a^{2}+8bc}}+\frac{b}{\sqrt{b^{2}+8ac}}+\frac{b}{\sqrt{b^{2}+8ac}}\geq \frac{1}{\sqrt{a^{3}+b^{3}+c^{3}+24abc}}$
but :
$1=(a+b+c)^{3}=a^{3}+b^{3}+c^{3}+6abc+3(a^{2}b+a^{2}c+b^{2}a+b^{2}c+c^{2}a+c^{2}b) \geq a^{3}+b^{3}+c^{3}+24abc$

, so indeed $\frac{1}{\sqrt{a^{3}+b^{3}+c^{3}+24abc}}\geq 1$ .

I don't exactly understand how you are applying Jensen's.

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HTA

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HTA

#6 h Apr 22, 2007, 6:03 AM • 11 Y

Y by Binomial-theorem, biomathematics, me9hanics, Adventure10, megarnie, son7, Vladimir_Djurica, Mango247, sabkx, ehuseyinyigit, and 1 other user

i think the problem can be solved by using UCT
by using UCT ,we need to prove that
$\frac{a}{\sqrt{a^{2}+8bc}}\geq \frac{a^{\frac{4}{3}}}{a^{\frac{4}{3}}+b^{\frac{4}{3}}+c^{\frac{4}{3}}}$
it follows that $(2a^{\frac{4}{3}}+b^{\frac{4}{3}}+c^{\frac{4}{3}})(b^{\frac{4}{3}}+c^{\frac{4}{3}}) \geq 8a^{\frac{2}{3}bc}$
writing two similar inequalities of b and c and adding them up , we obtain the desire result

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N.T.TUAN

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N.T.TUAN

#7 h Apr 22, 2007, 6:11 AM • 11 Y

Y by Trafalgar0246, ValidName, Polynom_Efendi, arqady, Adventure10, megarnie, son7, Vladimir_Djurica, Mango247, and 2 other users

@Phelpedo: $\text{LHS}=af(a^{2}+8bc)+bf(b^{2}+8ca)+cf(c^{2}+8ab)$ , that is all because $a+b+c=1$ .

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othman

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othman

#8 h Apr 22, 2007, 5:19 PM • 7 Y

Y by biomathematics, me9hanics, Adventure10, son7, Mango247, and 2 other users

i have an other idea : we can put, $e^{x}=\frac{bc}{a^{2}}$ , $e^{y}=\frac{ac}{b^{2}}$ and $e^{z}=\frac{ab}{c^{2}}$ to get the inequality we have only to apply Jensen to the function $f(x)=\frac{1}{1+e^{x}}$ with $\lambda_{1}=\lambda_{2}=\lambda_{3}=\frac{1}{3}$ but i'm not sure of the convexity of $f$

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Phelpedo

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Phelpedo

#9 h Apr 23, 2007, 1:07 AM • 8 Y

Y by YadisBeles, me9hanics, Adventure10, son7, Mango247, and 3 other users

HTA wrote:

i think the problem can be solved by using UCT
by using UCT ,we need to prove that
$\frac{a}{\sqrt{a^{2}+8bc}}\geq \frac{a^{\frac{4}{3}}}{a^{\frac{4}{3}}+b^{\frac{4}{3}}+c^{\frac{4}{3}}}$
it follows that $(2a^{\frac{4}{3}}+b^{\frac{4}{3}}+c^{\frac{4}{3}})(b^{\frac{4}{3}}+c^{\frac{4}{3}}) \geq 8a^{\frac{2}{3}bc}$
writing two similar inequalities of b and c and adding them up , we obtain the desire result

What does UCT stand for?

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othman

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othman

#10 h Apr 23, 2007, 4:14 PM • 4 Y

Y by Adventure10, Mango247, and 2 other users

i'have made a mistake th efunction $f(x)=\frac{1}{1+8e^{x}}$

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Nukular

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Nukular

#11 h Jun 1, 2007, 5:33 AM • 6 Y

Y by pavel kozlov, Adventure10, son7, Mango247, and 2 other users

That function $\frac{1}{1+8e^{x}}$ is only convex when $e^{x}> 1/4$ . However, things being less than 1/4 does not give the inequality much room to escape!

EDIT: If we have e^x <= 1/4 , then we can apply Jensen to the remaining two variables to obtain a one-variable inequality, which then dies to taking a derivative. So, this method *does* work, however ugly it is.

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epitomy01

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epitomy01

#12 h Feb 3, 2008, 8:46 AM • 8 Y

Y by JuanOrtiz, me9hanics, N0_NAME, Adventure10, son7, LLL2019, and 2 other users

Another solution:
Let $x = \sqrt \frac {a^2}{a^2 + 8bc}, y = \sqrt \frac {b^2}{b^2 + 8ac}, z = \sqrt \frac {c^2}{c^2 + 8ab}$ .
clearly, $\frac {1}{x^2} = 1 + \frac {8bc}{a^2}$ , so $( \frac {1}{x^2} - 1 )( \frac {1}{y^2} - 1) (\frac {1}{z^2} - 1) = 512$ . We wish to prove that $x + y + z \geq 1$ .
Let's prove that when $x + y + z = 1$ , then $( \frac {1}{x^2} - 1 )( \frac {1}{y^2} - 1) (\frac {1}{z^2} - 1) \geq 512$ . This is easy: $( \frac {1}{x^2} - 1 )( \frac {1}{y^2} - 1) (\frac {1}{z^2} - 1) = \frac { (1 - x)(1 + x)(1 - y)(1 + y)(1 - z)(1 + z) }{x^2 y^2 z^2} = \frac { (x + y)(y + z)(x + z)(2x + y + z)(2y + x + z)(2z + x + y)}{x^2 y^2 z^2} \geq \frac { 512 \sqrt {xy} \sqrt {yz} \sqrt {xz} \sqrt [4]{x^2 yz} \sqrt [4]{y^2 xz} \sqrt [4]{z^2 xy} }{x^2 y^2 z^2} = 512$ , as required.

This is sufficient, since for some $a,b,c$ , if $( \frac {1}{a^2} - 1 )( \frac {1}{b^2} - 1) (\frac {1}{c^2} - 1) = 512$ ... (*), letting $\lambda a = x, \lambda b = y, \lambda c = z$ , where $x + y + z = 1$ , then we have, $( \frac {1}{x^2} - 1 )( \frac {1}{y^2} - 1) (\frac {1}{z^2} - 1) \geq 512$ - then comparing with (*), we have $\lambda \leq 1$ , so $a + b + c \geq 1$ .

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dave05

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dave05

#13 h Feb 4, 2008, 3:46 PM • 13 Y

Y by hwl0304, theoryofuniverse, me9hanics, BobaFett101, shivprateek, Vladimir_Djurica, raigeki173, sabkx, ehuseyinyigit, Adventure10, Mango247, and 2 other users

Here's another solution, quite a simple one.
by Cauchy-Schwarz:
$(\frac{a}{\sqrt{a^2 + 8bc}} + \frac{b}{\sqrt{b^2 + 8ca}} + \frac{c}{\sqrt{c^2 + 8ab}})^2 (\sum_{cyc}{a(a^2+8bc)})$
$\geq(a+b+c)^3$
And:
$(a+b+c)^3\geq\sum_{cyc}{a(a^2+8bc)}$
Thus, we get:

$(\frac{a}{\sqrt{a^2 + 8bc}} + \frac{b}{\sqrt{b^2 + 8ca}} + \frac{c}{\sqrt{c^2 + 8ab}})^2$
$\geq\frac{(a+b+c)^3}{(\sum{a(a^2+8bc)})}$
$\geq1$
$Q.E.D.$

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crazyfehmy

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crazyfehmy

#14 h Jun 8, 2008, 9:54 PM • 9 Y

Y by me9hanics, Polynom_Efendi, Adventure10, Muaaz.SY, ZHEKSHEN, Chokechoke, ehuseyinyigit, Mango247, and 1 other user

Because of the homogeneity we can take $a.b.c = 1$ . Inequality equivalent to $\sqrt {\frac {a^3}{a^3 + 8}} + \sqrt {\frac {b^3}{b^3 + 8}} + \sqrt {\frac {c^3}{c^3 + 8}}\geq1$ . Let $a = \frac {2}{x} , b = \frac {2}{y} , c = \frac {2}{z}$ . $\frac {1}{\sqrt {1 + x^3}} + \frac {1}{\sqrt {1 + y^3}} + \frac {1}{\sqrt {1 + z^3}}\geq1$
By the Arithmetic Geometric Mean Inequality $\sqrt {(x + 1)(x^2 - x + 1)}\leq\frac {x^2 + 2}{2}$ and we will prove that $\frac {2}{x^2 + 2} + \frac {2}{y^2 + 2} + \frac {2}{z^2 + 2}\geq1$ $\Longleftrightarrow$ $2(x^2y^2 + y^2z^2 + x^2z^2) + 8(x^2 + y^2 + z^2) + 24\geq2(x^2y^2 + y^2z^2 + x^2z^2) + 4(x^2 + y^2 + z^2) + x^2y^2z^2$ $\Longleftrightarrow$ $x^2 + y^2 + z^2\geq12$ which is obviously true. ( $x.y.z=8$ )

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dgreenb801

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dgreenb801

#15 h Jun 24, 2008, 10:43 PM • 3 Y

Y by me9hanics, Adventure10, and 1 other user

What does UCT stand for? If it means showing that each of three fractions is greater the f(a)/(f(a)+f(b)+f(c)), hence their sum must be greater then 1, then what is the motivation for choosing f(x)= $x^{4/3}$ ?

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gracemoon124

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gracemoon124

#131 h Dec 7, 2023, 5:29 AM

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We have the following by Holder's:
$\left(\sum_{cyc}\frac{a}{\sqrt{a^2+8bc}}\right)^2\left(\sum_{cyc}a(a^2+8bc)\right)\ge (a+b+c)^3$ so it suffices to prove
$(a+b+c)^3\ge \sum_{cyc}a(a^2+8bc)$ equivalently
$a^2b+ab^2+bc^2+b^2c+c^2a+ca^2\ge 6abc$ which is true by direct AM-GM. $\square$

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Bryan0224

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Bryan0224

#133 h Aug 23, 2024, 1:43 AM

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ehuseyinyigit wrote:

Generalization 1
Let $a,b,c,p,\lambda$ be positive reals. Then prove that

$\dfrac{a^p}{\sqrt{a^{2p}+\lambda bc}}+\dfrac{b^p}{\sqrt{b^{2p}+\lambda ca}}+\dfrac{c^p}{\sqrt{c^{2p}+\lambda ab}}\geq \dfrac{\lambda ^2}{4\left(3\sqrt[3]{\lambda ^2}+4\right)}$

Tangent line or holder

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eg4334

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eg4334

#134 h Sep 4, 2024, 5:41 PM

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??

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ehuseyinyigit

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ehuseyinyigit

#135 h Sep 4, 2024, 10:07 PM

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Bryan0224 wrote:

Tangent line or holder

Yes, it has a proof by Holder's Inequality.

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ehuseyinyigit

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ehuseyinyigit

#136 h Sep 4, 2024, 10:08 PM

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Generalization 2
For any $a,b,c,\lambda$ ( $\lambda \geq 8$ ) positive reals, the inequality

$\dfrac{a}{\sqrt{a^{2}+\lambda bc}}+\dfrac{b}{\sqrt{b^{2}+\lambda ca}}+\dfrac{c}{\sqrt{c^{2}+\lambda ab}}\geq 1-\dfrac{\left(\sqrt[3]{\lambda ^2}-4\right)\left(\sqrt[3]{\lambda^2}+2\right)^2}{6\sqrt[3]{\lambda}\left(\lambda +2\sqrt[3]{\lambda}\right)+8}$
holds.

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anhduy98

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anhduy98

#137 h Sep 4, 2024, 11:10 PM

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ehuseyinyigit wrote:

Generalization 2
For any $a,b,c,\lambda$ ( $\lambda \geq 8$ ) positive reals, the inequality

$\dfrac{a}{\sqrt{a^{2}+\lambda bc}}+\dfrac{b}{\sqrt{b^{2}+\lambda ca}}+\dfrac{c}{\sqrt{c^{2}+\lambda ab}}\geq 1-\dfrac{\left(\sqrt[3]{\lambda ^2}-4\right)\left(\sqrt[3]{\lambda^2}+2\right)^2}{6\sqrt[3]{\lambda}\left(\lambda +2\sqrt[3]{\lambda}\right)+8}$
holds.

Because : $\frac{3}{\sqrt{1+\lambda}}\ge1-\dfrac{\left(\sqrt[3]{\lambda ^2}-4\right)\left(\sqrt[3]{\lambda^2}+2\right)^2}{6\sqrt[3]{\lambda}\left(\lambda +2\sqrt[3]{\lambda}\right)+8},\left(\lambda \geq 8\right).$

.

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ehuseyinyigit

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ehuseyinyigit

#138 h Sep 4, 2024, 11:49 PM • 1 Y

Y by MinhDangVN

Nice! Here was the $n$ -var extension (GENERALIZATION 3) :

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Aiden-1089

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Aiden-1089

#139 h Sep 5, 2024, 1:30 AM

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By Holder,
$\left( \sum_{cyc} \frac{a}{\sqrt{a^2 + 8bc}} \right)^2 \left( \sum_{cyc} a(a^2+8bc) \right) \geq (a+b+c)^3$ so it suffices to show that $a^3+b^3+c^3+24abc \leq (a+b+c)^3$ , or $\sum_{sym} a^2b \geq 6abc$ , which is trivial by Muirhead.

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fasttrust_12-mn

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fasttrust_12-mn

#140 h Sep 20, 2024, 10:06 AM

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huh....
$\left( \sum_{cyc} \frac{a}{\sqrt{a^2 + 8bc}} \right)^2 \left( \sum_{cyc} a(a^2+8bc) \right) \geq (a+b+c)^3$ Lemma: $(a+b+c)^3\geq a^3+b^3+c^3+24abc$
Proof:
$(a+b+c)^3=a^3+b^3+c^3+3\left(\sum_{sym} a^2b\right) +6abc \ge24abc+a^3+b^3+c^3\ \iff \sum_{sym} a^2b\ge 6abc$ which is true by AM-GM and we are done $\quad\square$

This post has been edited 2 times. Last edited by fasttrust_12-mn, Sep 20, 2024, 10:11 AM
Reason: typosus

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dimi07

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dimi07

#144 h Sep 22, 2024, 4:23 PM

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My first ever problem to post in this platform!

Okay so we have to show that

$\sum_{\text{cyc}} \frac{a}{\sqrt{a^2 + 8bc}} \geq 1.$
We can easily prove this by applying Hölder's inequality carefully. Let us choose $p = 2$ and $q =1$ . Applying Hölder's gives us:

$\sum_{\text{cyc}} \left( \frac{a}{\sqrt{a^2 + 8bc}} \right)^2 \cdot (a(a^2 + 8bc) \geq (a+b+c)^3.$
If we expand this, we get:

$\sum_{\text{cyc}} \frac{a^2}{a^2 + 8bc} \geq \frac{(a+b+c)^3}{\sum_{\text{cyc}} a(a^2 + 8bc)}.$
Thus, it suffices to show that

$\frac{(a+b+c)^3}{\sum_{\text{cyc}} a(a^2 + 8bc)} \geq 1,$
which is equivalent to proving

$(a+b+c)^3 \geq \sum_{\text{cyc}} a(a^2 + 8bc).$
Now, we compute

$\sum_{\text{cyc}} a(a^2 + 8bc) = \sum_{\text{cyc}} (a^3 + 8abc) = a^3 + b^3 + c^3 + 24abc.$
So we need to show that

$(a+b+c)^3 \geq a^3 + b^3 + c^3 + 24abc.$
Using the expansion, we have:

$(a+b+c)^3 = a^3 + b^3 + c^3 + 3(a+b)(b+c)(c+a).$
It suffices to prove that

$3(a+b)(b+c)(c+a) \geq 24abc.$
By the AM-GM inequality, we can derive:

$a^2b + a^2c + ab^2 + ac^2 + b^2c + bc^2 \geq 6abc.$
We can show this using AM-GM as follows:

1. For $a^2b + b^2c + c^2a$ :

$a^2b + b^2c + c^2a \geq 3 \sqrt[3]{(a^2b)(b^2c)(c^2a)} = 3abc.$
2. For $ab^2 + bc^2 + ca^2$ :

$ab^2 + bc^2 + ca^2 \geq 3 \sqrt[3]{(ab^2)(bc^2)(ca^2)} = 3abc.$
Adding these inequalities gives:

$a^2b + a^2c + ab^2 + ac^2 + b^2c + bc^2 \geq 6abc.$
Thus, we conclude that

$\sum_{\text{cyc}} \frac{a}{\sqrt{a^2 + 8bc}} \geq 1,$
and the proof is complete.

This post has been edited 1 time. Last edited by dimi07, Sep 23, 2024, 3:50 PM
Reason: Sorry p should be 1 since then the inequality cant be solved

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AshAuktober

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AshAuktober

#145 h Sep 26, 2024, 11:52 AM

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Claim: $\sum_{cyc} a(a^2 + 8bc) \le (a+b+c)^3$ .
Proof: Expand to get Muirhead's inequality. $\square$

Now, by Holder's inequality,
$\left(\frac{a}{\sqrt{a^2 + 8bc}} + \frac{b}{\sqrt{b^2 + 8ca}} + \frac{c}{\sqrt{c^2 + 8ab}}\right)^2 \left((a+b+c\right)^3)$ $\ge \left(\frac{a}{\sqrt{a^2 + 8bc}} + \frac{b}{\sqrt{b^2 + 8ca}} + \frac{c}{\sqrt{c^2 + 8ab}}\right)^2 \left(\sum_{cyc} a(a^2 + 8bc) \right)$ $\ge \left(\sum_{cyc} {a}\right)^3,$ so we're done.

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little-fermat

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little-fermat

#146 h Sep 26, 2024, 3:12 PM

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I have discussed this problem as an application of Holder in my inequality tutorial playlist. Here is the video

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EaZ_Shadow

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EaZ_Shadow

#147 h Dec 12, 2024, 11:44 PM • 1 Y

Y by WHO_LET_ME_COOK

It’s pretty simple. Just apply Cauchy schwarz then am gm kills the problem

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Maximilian113

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Maximilian113

#148 h Dec 13, 2024, 12:30 AM

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By Holder's Inequality, $\text{LHS}^2 \cdot \left( \sum_{cyc} a(a^2+8bc) \right) \geq (a+b+c)^3,$ so it suffices to show that $a^3+b^3+c^3+24abc \leq (a+b+c)^3$ which is well-known. (for a complete proof though, note we can fully expand everything and it will reduce to Muirhead.) QED

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Levieee

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Levieee

#149 h Friday at 9:08 PM

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there's a beautiful solution with AM GM idk why noone mentioned it
prove that:
$\frac{a}{\sqrt{a^2 + 8bc}}$ $\geq$ $\frac{a^{4/3}}{a^{4/3}+b^{4/3}+c^{4/3}}$
$\text{u can prove this buy squaring the denominator in the term that i wrote and applying AM GM on it}$ :O
(sorry for the handwritten solution)

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