IMO Shortlist 2011, Algebra 7

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orl

3647 posts

orl

#1 h Jul 11, 2012, 10:24 PM • 12 Y

Y by tastymath75025, tritanngo99, Adventure10, Mango247, Rounak_iitr, ehuseyinyigit, cubres, and 5 other users

Let $a,b$ and $c$ be positive real numbers satisfying $\min(a+b,b+c,c+a) > \sqrt{2}$ and $a^2+b^2+c^2=3.$ Prove that

$\frac{a}{(b+c-a)^2} + \frac{b}{(c+a-b)^2} + \frac{c}{(a+b-c)^2} \geq \frac{3}{(abc)^2}.$

Proposed by Titu Andreescu, Saudi Arabia

This post has been edited 2 times. Last edited by orl, Jul 24, 2012, 1:35 PM

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oneplusone

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oneplusone

#2 h Jul 12, 2012, 5:28 AM • 35 Y

Y by Nguyenhuyhoang, huyvietnam, sayantanchakraborty, MathPanda1, AdBondEvent, quangminhltv99, JasperL, AlgebraFC, yiwen, Wizard_32, abdelkrim, HolyMath, karitoshi, Wizard0001, srijonrick, megarnie, myh2910, ThisNameIsNotAvailable, qwedsazxc, WinterSecret, Adventure10, Mango247, ehuseyinyigit, Rayanelba, farhad.fritl, and 10 other users

It is easy to check that $a+b-c>0$ and its cyclic counterparts. Then by Holder's,
$\sum_{cyc}{\frac{a}{(b+c-a)^2}}\sum_{cyc}a^2(b+c-a)\sum_{cyc}a^3(b+c-a)\geq (\sum_{cyc}a^2)^3=27$
By Schurs,
$\sum_{cyc}a^2(b+c-a)\leq 3abc$ and
$\sum_{cyc}a^3(b+c-a)\leq abc(a+b+c)$ Therefore
$\sum_{cyc}{\frac{a}{(b+c-a)^2}}\geq \frac{9}{(abc)^2(a+b+c)}\geq\frac{3}{(abc)^2}$

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daniel73

253 posts

daniel73

#3 h Jul 12, 2012, 7:24 AM • 7 Y

Y by yshk, Adventure10, Mango247, and 4 other users

For those who do not quickly see why $b+c-a>0$ :

Click to reveal hidden text

Alternative solution without Holder's inequality:

Click to reveal hidden text

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mathbuzz

803 posts

mathbuzz

#4 h Jul 13, 2012, 3:58 AM • 4 Y

Y by ehuseyinyigit, Adventure10, Mango247, and 1 other user

for solving it , we first prove a claim
claim--- let x,y,z be 3 positive reals. they satisfy $min(x+y,y+z,z+x)> 2^{1/2}$ and $x^2+y^2+z^2=3$ .
then x,y,z are the sides of a triangle.
proof--- here we clearly assume an ordering $x >= y >= z$ .
then clearly min(x+y,y+z,z+x)=y+z. then y+z > $2^{1/2}$ . then 2 $(y^2+z^2)$ >= $(z+y)^2$ . so, $y^2+z^2$ > $1$ . now , clearly , $y+z >x$ [if not , assume that $y+z <= x$ . then $x >= y+z > 2^{1/2}$ . so, $x^2 >2$ , which contradicts $x^2+y^2+z^2$ =3]. so , x,y,z are the sides of a triangle.

now , applying it to our problem , we get that , a,b,c are the sides of a triangle.so , put a=p+q , b=q+r , c=r+p.
then the given condition translates into the constraint
$2(p^2+q^2+r^2+pq+qr+rp)$ =3 and then applying LM method is sufficient . [here we must keep in mind the condition min(p+q,q+r,r+p) > $sqrt.2$ for choosing
the values of p,q,r for minimizing the expression (if required)]. the minimum value is at p=q=r=1/2 [calculation is tedious!]

Click to reveal hidden text

This post has been edited 1 time. Last edited by mathbuzz, Aug 13, 2012, 12:15 PM

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mahanmath

1354 posts

mahanmath

#5 h Jul 13, 2012, 11:59 PM • 8 Y

Y by bvdsf, Adventure10, Mango247, and 5 other users

We had got this inequality at an exam

, This is what I wrote that time :
Click to reveal hidden text

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Anzoteh

126 posts

Anzoteh

#6 h Dec 4, 2012, 1:51 AM • 4 Y

Y by Adventure10, Mango247, and 2 other users

Here is mine. (Outline)
1. Show that $a, b, c$ are sides of triangle, so that $\exists x, y, z\in\mathbb R^{+}$ with $a=x+y, b=x+z, c=y+z$ (which is shown by someone else).
2. The inequality becomes $[x^{2}y^{2}(x+y)+y^{2}z^{2}(y+z)+z^{2}x^{2}(z+x)]\ge 12x^{2}y^{2}z^{2}$
3. Show that $x+y+z\le\frac{3}{2},$ and homogenize the expression in (2) by multiplying LHS by $\frac{2}{3} (x+y+z)$ and RHS by $\frac {1}{27} ((x+y)^{2}+(y+z)^{2}+(z+x)^{2})^{3}.$
4. Expand (3) and we only need to use Muirhead to settle everything (See below)!

So it would be: (all symmetric sum)
$(3\sum x^{8}y^{4}z^{0}+6\sum x^{8}y^{3}z^{1}+3\sum x^{8}y^{2}z^{2}+12\sum x^{7}y^{5}z^{0}+33\sum x^{7}y^{4}z^{1}+39\sum x^{7}y^{3}z^{2}+9\sum x^{6}y^{6}z^{0}+69\sum x^{6}y^{5}z^{1}+102\sum x^{6}y^{4}z^{2}+51\sum x^{6}y^{3}z^{3}+69\sum x^{5}y^{5}z^{2}+153\sum x^{5}y^{4}z^{3}+27\sum x^{4}y^{4}z^{4})\ge (8\sum x^{8}y^{2}z^{2}+48\sum x^{7}y^{3}z^{2}+96\sum x^{6}y^{4}z^{2}+72\sum x^{6}y^{3}z^{3}+56\sum x^{5}y^{5}z^{2}+240\sum x^{5}y^{4}z^{3}+56\sum x^{4}y^{4}z^{4})$
which is obvious by Muirhead

It would be incredible that an A7 problem can be solved by expanding alone.

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utkarshgupta

2280 posts

utkarshgupta

#7 h Sep 10, 2014, 10:20 AM • 2 Y

Y by Adventure10 and 1 other user

Here will
$\prod (a+b-c) \ge 1$ ?

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Abubakir

68 posts

Abubakir

#8 h Sep 2, 2015, 1:53 PM • 2 Y

Y by Adventure10, Mango247

mathbuzz wrote:

for solving it , we first prove a claim
claim--- let x,y,z be 3 positive reals. they satisfy $min(x+y,y+z,z+x)> 2^{1/2}$ and $x^2+y^2+z^2=3$ .
then x,y,z are the sides of a triangle.
proof--- here we clearly assume an ordering $x >= y >= z$ .
then clearly min(x+y,y+z,z+x)=y+z. then y+z > $2^{1/2}$ . then 2 $(y^2+z^2)$ >= $(z+y)^2$ . so, $y^2+z^2$ > $1$ . now , clearly , $y+z >x$ [if not , assume that $y+z <= x$ . then $x >= y+z > 2^{1/2}$ . so, $x^2 >2$ , which contradicts $x^2+y^2+z^2$ =3]. so , x,y,z are the sides of a triangle.

now , applying it to our problem , we get that , a,b,c are the sides of a triangle.so , put a=p+q , b=q+r , c=r+p.
then the given condition translates into the constraint
$2(p^2+q^2+r^2+pq+qr+rp)$ =3 and then applying LM method is sufficient . [here we must keep in mind the condition min(p+q,q+r,r+p) > $sqrt.2$ for choosing
the values of p,q,r for minimizing the expression (if required)]. the minimum value is at p=q=r=1/2 [calculation is tedious!]

Click to reveal hidden text

What does abbreviation "LM" mean?

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vjdjmathaddict

502 posts

vjdjmathaddict

#9 h Apr 18, 2017, 3:16 PM • 1 Y

Y by Adventure10

lagrange multiplier

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ahmedAbd

91 posts

ahmedAbd

#10 h Apr 18, 2017, 7:46 PM • 4 Y

Y by Wizard_32, kiyoras_2001, Adventure10, Mango247

What does Titu Andreescu have to do with Saudi Arabia?

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Medjl

757 posts

Medjl

#11 h Apr 18, 2017, 7:59 PM • 1 Y

Y by Adventure10

ahmedAbd wrote:

What does Titu Andreescu have to do with Saudi Arabia?

he was invited to teach Saudian team

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mathcool2009

352 posts

mathcool2009

#12 h Apr 22, 2018, 1:24 AM • 12 Y

Y by 62861, MathStudent2002, Makorn, niyu, sunfishho, aops29, yayups, myh2910, Adventure10, HamstPan38825, Mango247, Quidditch

I wasn't expecting this to be possible.
Degree 14 Polynomial Bash

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H.HAFEZI2000

328 posts

H.HAFEZI2000

#13 h Jul 14, 2018, 7:49 PM • 3 Y

Y by 554183, Adventure10, Mango247

ahmedAbd wrote:

What does Titu Andreescu have to do with Saudi Arabia?

they probably paid for this

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Taha1381

816 posts

Taha1381

#14 h Oct 27, 2018, 2:37 PM • 1 Y

Y by Adventure10

oneplusone wrote:

It is easy to check that $a+b-c>0$ and its cyclic counterparts. Then by Holder's,
$\sum_{cyc}{\frac{a}{(b+c-a)^2}}\sum_{cyc}a^2(b+c-a)\sum_{cyc}a^3(b+c-a)\geq (\sum_{cyc}a^2)^3=27$

What was the motivation for this kind of Holder usage?

This post has been edited 2 times. Last edited by Taha1381, Oct 27, 2018, 2:38 PM

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WolfusA

1900 posts

WolfusA

#15 h Aug 1, 2019, 12:10 PM • 8 Y

Y by Sugiyem, PickleSauce, 508669, L567, mathscrazy, ehuseyinyigit, Adventure10, farhad.fritl

H.HAFEZI2000 wrote:

ahmedAbd wrote:

What does Titu Andreescu have to do with Saudi Arabia?

they probably paid for this

Ten oil barrels for sure.

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Math-wiz

6107 posts

Math-wiz

#16 h Nov 9, 2019, 4:50 AM • 1 Y

Y by Adventure10

orl wrote:

Let $a,b$ and $c$ be positive real numbers satisfying $\min(a+b,b+c,c+a) > \sqrt{2}$ and $a^2+b^2+c^2=3.$ Prove that

$\frac{a}{(b+c-a)^2} + \frac{b}{(c+a-b)^2} + \frac{c}{(a+b-c)^2} \geq \frac{3}{(abc)^2}.$
Proposed by Titu Andreescu, Saudi Arabia

Everything is fine, but why Titu Andreescu, Saudi Arabia! Sorry if this is considered spam

This post has been edited 1 time. Last edited by Math-wiz, Nov 9, 2019, 4:50 AM

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solver1104

510 posts

solver1104

#17 h Nov 9, 2019, 5:49 AM • 2 Y

Y by Adventure10, Mango247

H.HAFEZI2000 wrote:

ahmedAbd wrote:

What does Titu Andreescu have to do with Saudi Arabia?

they probably paid for this

did you not see this

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Math-wiz

6107 posts

Math-wiz

#18 h Nov 9, 2019, 5:54 AM • 1 Y

Y by Adventure10

solver1104 wrote:

H.HAFEZI2000 wrote:

ahmedAbd wrote:

What does Titu Andreescu have to do with Saudi Arabia?

they probably paid for this

did you not see this

Ohh, sorry

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ZETA_in_olympiad

2211 posts

ZETA_in_olympiad

#20 h Jul 1, 2022, 10:40 AM

Y by

Note that $a+b-c, b+c-a$ and $a+c-b$ are positive. Thus by Hölder: $\sum_{\text{cyc}} a^3(b+c-a) \sum_{\text{cyc}} a^2(b+c-a) \sum_{\text{cyc}} \frac{a}{(b+c-a)^2} \geq 27.$ Since by Schur: $3ab\geq \sum_{\text{cyc}} a^2(b+c-a) \quad \text{and} \quad abc(a+b+c)\geq \sum_{\text{cyc}} a^3(b+c-a).$ Thus the inequality of the problem holds.

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strong_boy

261 posts

strong_boy

#21 h Nov 23, 2022, 12:06 AM • 2 Y

Y by Mango247, Mango247

It is easy to see $\sum \frac{a}{(b+c-a)^2} = \sum \frac{a^6}{a^5(b+c-a)^2}$ . Now we can use $T-2$ lemma :

$\sum \frac{a^6}{a^5(b+c-a)^2} = \sum \frac{a^6}{(a^{\frac{5}{2}}(b+c-a))^2} \geq \frac{27}{(\sum a^{\frac{5}{2}}(b+c-a))^2}$
Now by schur it is easy to see :

$\sum a^{\frac{5}{2}}(b+c-a))^2 \leq abc(\sum \sqrt{a})$
Now we only need to prove :
$\frac{27}{abc(\sum \sqrt{a})^2} \geq \frac{3}{(abc)^2}$
Now we need to prove $\sum \sqrt{a} \leq \sqrt{3}$ .And it is easy by Holder .

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awesomeming327.

1699 posts

awesomeming327.

#22 h Jun 3, 2023, 12:13 AM

Y by

Let $S$ denote the left hand side. We know that by Schur's Inequality,
$\begin{align*} a^{1.5}(a-b)(a-c)+b^{1.5}(b-c)(b-a)+c^{1.5}(c-a)(c-b)\ge 0\\ a^{1.5}bc+b^{1.5}ca+c^{1.5}ab\ge a^{2.5}(b+c-a) + b^{2.5}(c+a-b)+c^{2.5}(a+b-c) \end{align*}$ Since $(\sqrt{a}+\sqrt{b}+\sqrt{c})^4\le 27(a^2+b^2+c^2)=81$ by Power Mean Inequality, we have
$3abc \ge a^{2.5}(b+c-a) + b^{2.5}(c+a-b)+c^{2.5}(a+b-c)$ By Holder's Inequality, $S(3abc)^2 \ge (a^2+b^2+c^2)^3=27$ so the desired holds.

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sqing

41782 posts

sqing

#23 h Jun 3, 2023, 7:49 AM

Y by

Let $a,b$ and $c$ be positive real numbers satisfying $\min(a+b,b+c,c+a) > \sqrt{2}$ and $a^2+b^2+c^2=3.$ Prove or disprove
$\frac{a}{(b+c-a)^3} + \frac{b}{(c+a-b)^3} + \frac{c}{(a+b-c)^3} \geq \frac{3}{(abc)^3}$

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S.Das93

707 posts

S.Das93

#24 h Jun 3, 2023, 3:05 PM

Y by

Can we solve this in a similar way by showing the Triangle Inequality and then substituting the respective $a=u+v,b=v+w,c=w+u$ ?

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bin_sherlo

707 posts

bin_sherlo

#25 h Apr 15, 2025, 7:34 PM

Y by

Since $2(a^2+b^2)\geq (a+b)^2>2$ we get $a+b>\sqrt 2>\sqrt{3-a^2-b^2}=c$ thus, $a,b,c$ satisfy the triangle inequality. Let $a=\frac{y+z}{2},b=\frac{x+z}{2},c=\frac{x+y}{2}$ .
$\sum{x}\sum{\frac{1}{x^2}}-\sum{\frac{1}{x}}=\sum{\frac{y+z}{x^2}}\overset{?}{\geq} \frac{384}{(y+z)^2(x+z)^2(x+y)^2}$ Let $x+y+z=3u, \ xy+yz+zx=3v^2,\ xyz=w^3$ where $u^2+v^2=2$ .
$3u.\frac{9v^4-6uw^3}{w^6}-\frac{3v^2}{w^3}\overset{?}{\geq} \frac{384}{(9uv^2-w^3)^2}\iff \frac{u(9v^4-6uw^3)}{w^6}\overset{?}{\geq} \frac{v^2}{w^3}+\frac{128}{(9uv^2-w^3)^2}$ Note that $v^4\geq uw^3$ holds and $u\geq 1$ since $2=u^2+v^2\leq 2u^2$ .
$\frac{u(9v^4-6uw^3)}{w^6}\geq \frac{3uv^4}{w^6}\geq \frac{3u^2}{w^3}\geq \frac{v^2}{w^3}+\frac{2u^2}{w^3}\geq \frac{v^2}{w^3}+\frac{2}{u^2v^4}\geq \frac{v^2}{w^3}+\frac{128}{(9uv^2-w^3)^2}$ As desired. $\blacksquare$

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