Come on, not this again. it has been posted many times before, I think. From the subtitutions $a=y+z, b=z+x, c=x+y$ and the inequality is equivalent with $\sum\frac{x^{2}}{y}\geq\sum x.$ From here it is clear how to proceed.

Hi Cezar Lupu! As I do this IMO again, I found the very interesting solution for it and it is only 1 line . Can u believe it? He he . Ok ! Here is mine ( Any please don't ask me why I found it!); IMO inequality is equivalent to this "obvious inequality": $\frac12((a-b)^2(a+b-c)(b+c-a)+(b-c)^2(b+c-a)(a+c-b)+(c-a)^2(a+c-b)(a+b-c))\geq 0$ Right !

Sure enough, that's what i did.I was wondering if you could use Schur to solve it? Anyone? Ashwath

Gibbenergy wrote: Hi Cezar Lupu! As I do this IMO again, I found the very interesting solution for it and it is only 1 line . Can u believe it? He he . Ok ! Here is mine ( Any please don't ask me why I found it!); IMO inequality is equivalent to this "obvious inequality": $\frac12((a-b)^2(a+b-c)(b+c-a)+(b-c)^2(b+c-a)(a+c-b)+(c-a)^2(a+c-b)(a+b-c))\geq 0$ Right ! It's not new and obviously it's not your solution, Gibb . The ineq is homogeneous , so i can easily change the ineq into the fomula: $A(b-c)^2+B(c-a)^2+C(a-b)^2 \geq 0$

Hey Gibb this one-line solution had been given by an IMO contestant.

Hey guys! That is my sol and I have found it by myself. I don't care who has posted it before. I found my sols because as I know the IMO US team at that time was also use the trick that put $x=u+v...$ It is obviously unnatural. So it is not important with the way they showed. The question is how can the author generate it? I believed that the author of IMO 83 used the same technique that I have discovered now to make it. So it doesn't matter that my sols now is posted before. If you say you can make $A (b-c)^2$ i believed it is the theory. The important thing is how can you find $A$? $(b-c)^2....$ and also their permutation. If you can show me that technique, I would admire you as my best teacher. I feel that I don't want to share any my technique to you guys either. I will keep it as my secrect. Thanks.

Gibbenergy wrote: Hey guys! That is my sol and I have found it by myself. I don't care who has posted it before. I found my sols because as I know the IMO US team at that time was also use the trick that put $x=u+v...$ It is obviously unnatural. So it is not important with the way they showed. The question is how can the author generate it? I believed that the author of IMO 83 used the same technique that I have discovered now to make it. So it doesn't matter that my sols now is posted before. If you say you can make $A (b-c)^2$ i believed it is the theory. The important thing is how can you find $A$? $(b-c)^2....$ and also their permutation. If you can show me that technique, I would admire you as my best teacher. I feel that I don't want to share any my technique to you guys either. I will keep it as my secrect. Thanks. Ha! It's really funny. If you are willing to search on the forum, you can see that many people here prove inequalities by rewriting the inequality into sum of squares. Obviously they have a technique, or at least a general way to write inequalities in that form. Although it may not be able to be written rigorously, a general approach does exist I think. Of course you have the right not to share your technique and keep it as your secret. But I just wonder, if you don't even want to share, why do you hope others are willing to share with you? It's just like you want others to share something with you while you are reluctant to hand out anything. What a joke.

What a ridiculous conversation. Well, there are plenty of people willing to share NEW and interesting techniques with us, so we shouldn't worry at all. And, oh yes, topic locked.

I use Ravi's substitution. Let $a=x+y, b=y+z, c=z+x$. Some computation give...: \[ x^3z+xy^3+yz^3\geq xyz(x+y+z), \] \[ \frac{x^2}{y}+\frac{y^2}{z}+\frac{z^2}{x}\geq x+y+z, \] what is true by rearrangement inequality. Equality holds, when $x=y=z$, so when $a=b=c$.

Also discussed at http://www.mathlinks.ro/Forum/viewtopic.php?t=55406 and http://www.mathlinks.ro/Forum/viewtopic.php?t=15558 (posts #8-#10) . darij

The inequality equivalent to: \[a^{3}b+b^{3}c+c^{3}a \ge a^{2}b^{2}+b^{2}c^{2}+c^{2}a^{2}\] And this is a result of Moorhead Inequality \[\left({3,1,0}\right) > \left({2,2,0}\right) \]

This is an old topic but I think that it's worth pointing out that the LHS of your inequality is cyclic, not symmetric. Muirhead only works on symmetric sums, so the rearrangement method is needed.

We can use the Entirely Mixing variable method to solve this problem (but this method is so strong to solve such an easy problem). It's sufficient to consider one case $a=b+c$ and the rest is easy.

GacKiem wrote: We can use the Entirely Mixing variable method to solve this problem (but this method is so strong to solve such an easy problem). It's sufficient to consider one case $ a = b + c$ and the rest is easy. Sorry about reviving an old thread, but after reading http://reflections.awesomemath.org/2006_5/2006_5_entirelymixing.pdf, I still don't get the method. I can't show that replacing $ (a,b,c)$ with $ (a-c,b-c,0)$ will necessarily decrease LHS.

oocal or whatever, You can prove your obtained inequality by Cauchy for the reals $ \dfrac{a}{sqrtc},....$ and $ sqrtc....$.

Aravind Srinivas L wrote: oocal or whatever, You can prove your obtained inequality by Cauchy for the reals $ \dfrac{a}{sqrtc},....$ and $ sqrtc....$.

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ashwath.rabindranath wrote: Let $ a$, $ b$ and $ c$ be the lengths of the sides of a triangle. Prove that \[ a^{2}b(a - b) + b^{2}c(b - c) + c^{2}a(c - a)\ge 0. \] Determine when equality occurs. This problem has many proofs. Here is the proof using a nice property of quadratic polynomial Assuming (WLOG) that $ a \ge c \ge b$, then using AM-GM Inequality, we have $ a^2 \ge 2ac-c^2$. It suffices to show that \[ (2ac-c^2)b(a-b) +b^2c(b-c)+c^2a(c-a) \ge 0\] \[ (2a-c)b(a-b) +b^2(b-c)+ca(c-a) \ge 0\] \[ f(a)=(2b-c)a^2+(c^2-bc-2b^2)a+b^3 \ge 0\] If $ 2b \ge c$, using $ a^2 \ge 2ac-c^2$ again, it remains to show that \[ (2b-c)(2ac-c^2) +(c^2-bc-2b^2)a+b^3 \ge 0\] \[ (2b-c)(c-b)a+b^3-2bc^2+c^3 \ge 0\] which is true because \[ LHS \ge (2b-c)(c-b)c+b^3-2bc^2+c^3=b(b-c)^2 \ge 0\] If $ c \ge 2b$, then $ f(a)$ is a quadratic polynomial with the highest coefficient is negative and we see that $ c \le a \le b+c$, it follows that \[ f(a) \ge \min\{f(c),f(b+c)\}\] But \[ f(c)=b(b-c)^2\] and \[ f(b+c)=b^3 \ge 0\] Thus, $ f(a) \ge 0$ and the proof is completed. P/s: I think that this proof is new and nice.

wlog $ a \geq b \geq c$ $ c(a+b-c) \geq b(c+a-b)$ and $ c(a+b-c)-b(c+a-b)=(b-c)(b+c-a) \geq 0$ ve $ b(b+c-a)-a(c+a-b)=(a-b)(c+a-b) \geq 0$ we obtained that $ c(a+b-c) \geq b(c+a-b) \geq a(a+b-c)$ $ a \geq b \geq c$ and $ \frac{1}{c}\geq\frac{1}{b}\geq\frac{1}{a}$ By the rearrangement inequality; $ \frac{1}{c}a(b+c-a)+\frac{1}{a}b(c+a-b)+\frac{1}{b}c(a+b-c)\leq \frac{1}{a}a(b+c-a)+\frac{1}{b}b(c+a-b)+\frac{1}{c}c(a+b-c)=a+b+c$ which is desired inequality

Here is another way: after the substitution a=x+y, etc. use weighted AM-GM with the weights 4/7, 1/7 and 2/7

Making a Ravi Substitution yields \begin{align*} \text{ LHS } &= (y+z)^{2}(x+z)((y+z) - (x+z)) + (x+z)^{2}(x+y)((x+z) - (x+y)) + (x+y)^{2}(y+z)((x+y) - (y+z)) \\ &=2(xy^3 + x^3z - x^2yz - xy^2z - xyz^2 + yz^3) \\ &\implies \text{ LHS } \Leftrightarrow xy^3 + x^3z - x^2yz - xy^2z - xyz^2 + yz^3 \geq 0\\ &\implies \text{ LHS } \Leftrightarrow xy^3 + x^3z + yz^3 \geq x^2yz+xy^2z+xyz^2.\\ \end{align*}This inequality is homogenous, therefore we let $xyz=1$ to get $$\text{ LHS } \Leftrightarrow \frac{y^2}{z} + \frac{x^2}{y} + \frac{z^2}{x} \geq x+y+z,$$which is true by the Rearrangement Inequality on sequences $(x^2,y^2,z^2)$ and $(\tfrac{1}{x},\tfrac{1}{y},\tfrac{1}{z}).$ Equality holds when either of these sequences are constant, or whenever $x=y=z \implies a=b=c.$ $\mathbb{Q.E.D.}$

Insta-killed by Ravi! Ravi sub, we get that we want to show that $\sum_{\text{sym}}x^3y\geq \sum_{\text{sym}}x^2yz,$ which is just Muirhead. Why does everyone overcomplicate these type of problems and use rearrangement and homogenizing when it's just a direct Muirhead application? I know that Mildorf said not to cite Muirhead, but I am pretty sure its well known enough by now(Mildorf wrote his article in 2005)

gold46 wrote: Muirhead and Rearrangement inequalities don't work here. Is this dude just clowning A few others also say that Muirhead doesn't work because the inequality is cyclic, not symmetric, but idek :shrug: JustKeepRunning wrote: Why does everyone overcomplicate these type of problems and use rearrangement and homogenizing when it's just a direct Muirhead application? I know that Mildorf said not to cite Muirhead, but I am pretty sure its well known enough by now(Mildorf wrote his article in 2005) Well homogenization was just a useless step because dividing by $xyz$ in my expression performs the same task. Tbh I have no idea why I used it, but either way it's automatic. For reference here is the Mildorf article. Also idk Muirhead that well and this is a direct rearrangement application too right?

@above what do you mean? The symemtric is just $2$ times the cyclic...

JustKeepRunning wrote: @above what do you mean? The symemtric is just $2$ times the cyclic... I don't know why - I just remarked that some of the other posts stated that rearrangement/muirhead don't work.

JustKeepRunning wrote: @above what do you mean? The symemtric is just $2$ times the cyclic... For instance $\sum_{\text{sym}}a^2b=a^2b+ab^2+b^2c+bc^2+c^2a+ca^2$ but $2\sum_{\text{cyc}}a^2b=2a^2b+2b^2c+2c^2a$. Making Muirhead work isn't trivial as far as I can tell.

With Ravi ($a=x+y,b=y+z,c=z+x$) we have to prove: $$(x+y)^2(y+z)(x-z)+(y+z)^2(z+x)(y-x)+(z+x)^2(x+y)(z-y)\ge0$$$$\Leftrightarrow2x^3z+2xy^3+2yz^3\ge2x^2yz+2xy^2z+2xyz^2,$$which is true since $\frac{x^2}y+\frac{y^2}z+\frac{z^2}x\ge x+y+z$ by C-S.

Since sides of triangle it is true if we WLOG, $$a\geq b\geq c \implies a^2+bc \geq b^2+ca\geq c^2+ab.$$By rearrangement inequality, \begin{align*} \sum_{\text{cyc}} \frac{a^2+bc}{c} \geq \sum_{\text{cyc}} \frac{a^2+bc}{a}\\ \implies \sum_{\text{cyc}} \frac{a^2}{c} \geq \sum_{\text{cyc}} \frac{bc}{a} \\ \implies \sum_{\text{cyc}} a^3b \geq \sum_{\text{cyc}} a^2b^2.\end{align*}Equality when $a=b=c.$ And the claim follows. $\square$

Ravi inequality seems cool, could I have a introductory note on it?

ZETA_in_olympiad wrote: Ravi inequality seems cool, could I have a introductory note on it? here

Rewrite the inequality as $$\sum_{cyc}a^2b(a-b)\geq 0.$$Let $a=y+z,b=x+z,c=x+y$. Then, this becomes $$\sum_{cyc}(y+z)^2(x+z)(y-x)\geq 0.$$Expanding, this is $$\sum_{cyc}(y^2+2yz+z^2)(xy-x^2+yz-xz)\geq 0$$$$\sum_{cyc}xy^3+y^3z+yz^3-xz^3-2x^2yz+xy^2z-xyz^2-x^2y^2-x^2z^2+2y^2z^2\geq 0.$$Note that $$\sum_{cyc}xy^2z-xyz^2=\sum_{cyc} -x^2y^2-x^2z^2+2y^2z^2=\sum_{cyc}y^3z-xz^3=0,$$so it suffices to show that $$\sum_{cyc}xy^3+yz^3-2x^2yz\geq 0.$$However, this is just $$\sum_{sym}x^3y\geq \sum_{sym} x^2yz,$$which is just Muirhead's since $(3,1,0)\succ (2,1,1)$ so we are done. Equality occurs when $x=y=z$ or equivalently $a=b=c$ from the Muirhead step.

ugh.. i just had to expand it bruh After my attachment, the inequality follows from C-S since $LHS(y+z+x)\ge (x+y+z)(y+z+x). \blacksquare$

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oops other sol less expanding but much harder to find

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abasse wrote: a²b(a-b)>=a²(a-c) (a-b) because b>=a-c the make the same thing We are a²b(a-b)+b²c(b-c)+c²a(c-a)>= a²(a-b) (a-c) +b²(b-a) (b-c) +c²(c-b)(c-a) >=0 schur but then what about the equality case, since b>a-c. I think if a-b is negative then "a²b(a-b)>=a²(a-c) (a-b)" is not correct

Ravi Sub: $$\sum_{\text{cyc}}(y+z)^2(x+z)(y-x) \geq 0$$Expand: $$\sum_{\text{cyc}} xy^3+xy^2z+y^3z+2y^2z^2+yz^3 \geq \sum_{\text{cyc}} x^2y^2 + 2x^2yz + x^2z^2 + xyz^2 + xz^3$$Cancel terms: $$\sum_{\text{cyc}} xy^3+yz^3 \geq \sum_{\text{cyc}} 2x^2yz$$or equivalently $$\sum_{\text{sym}} x^3y \geq \sum_{\text{sym}} x^2yz$$which by $\{ 3, 1, 0\} \succ \{ 2, 1, 1\}$ we are done. Equality holds when the triangle is equilateral.