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
Problem
Source: IMO Shortlist 2011, Algebra 7
Tags: inequalities, algebra, IMO Shortlist
12.07.2012 08:28
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 \[\sum_{cyc}{\frac{a}{(b+c-a)^2}}\geq \frac{9}{(abc)^2(a+b+c)}\geq\frac{3}{(abc)^2}\]
12.07.2012 10:24
For those who do not quickly see why $b+c-a>0$:
Alternative solution without Holder's inequality:
13.07.2012 06:58
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!]
14.07.2012 02:59
We had got this inequality at an exam , This is what I wrote that time :
04.12.2012 03:51
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.
10.09.2014 13:20
Here will $\prod (a+b-c) \ge 1$ ?
02.09.2015 16:53
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!]
What does abbreviation "LM" mean?
18.04.2017 18:16
lagrange multiplier
18.04.2017 22:46
What does Titu Andreescu have to do with Saudi Arabia?
18.04.2017 22:59
ahmedAbd wrote: What does Titu Andreescu have to do with Saudi Arabia? he was invited to teach Saudian team
22.04.2018 04:24
I wasn't expecting this to be possible.
14.07.2018 22:49
ahmedAbd wrote: What does Titu Andreescu have to do with Saudi Arabia? they probably paid for this
27.10.2018 17:37
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?
01.08.2019 15:10
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.
09.11.2019 07:50
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
09.11.2019 08:49
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
09.11.2019 08:54
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
01.07.2022 13:40
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.
23.11.2022 03:06
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 .
03.06.2023 03:13
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.
03.06.2023 10:49
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 $$ \frac{a}{(b+c-a)^3} + \frac{b}{(c+a-b)^3} + \frac{c}{(a+b-c)^3} \geq \frac{3}{(abc)^3}$$
03.06.2023 18:05
Can we solve this in a similar way by showing the Triangle Inequality and then substituting the respective $a=u+v,b=v+w,c=w+u$ ?