Hi zhaoli, Thanks for the problem, but do you know the other BMO problems?? If so, can you please post them. Regards, Zhivko

rewrite it to $\frac{c+a}{b}(a-b)^2+\frac{a+b+c}{c}(b-c)^2+\frac{a+b+c}{a}(c-a)^2 \geq 3(a-b)^2$

Rewrite the intial inequality to: $\frac{(a-b)^2}{b}+\frac{(b-c)^2}{c}+\frac{(c-a)^2}{a} \geq \frac{4(a-b)^2}{a+b+c}$. This is equivalent to $(a+b+c)\left(\frac{(a-b)^2}{b}+\frac{(b-c)^2}{c}+\frac{(c-a)^2}{a} \right) \geq 4(a-b)^2$. Using Cauchy one can prove $(a+b+c)\left(\frac{(a-b)^2}{b}+\frac{(b-c)^2}{c}+\frac{(c-a)^2}{a} \right) \geq 4$ $($max$(a,b,c)-$min$(a,b,c)$$)^2$. q.e.d.

Nice solution! I always don't know how to deal with some non-symmetric inequalities.

The problems of Balkan Mathematical Olympiad from 1984-2003 can be found here The problems of Balkan Mathematical Olympiad from 2004 can be found at: http://www.artofproblemsolving.com/Forum/topic-5591.html

Thanks zhaoli, but what I meant were the problems from BMO 2005 (you posted only the third one). Well anyways, they will certainly appear at some point. Regards, Zhivko

Jivko, look better, they are in the corresponding subforums.

Yes, harazi, but some of us are only looking in the algebra section ... and are expecting to find all the BMO 2005 problems there. Thanks, harazi! Zhivko

jivko777 wrote: Yes, harazi, but some of us are only looking in the algebra section ... and are expecting to find all the BMO 2005 problems there. Thanks, harazi! Zhivko Check out this thread: http://www.mathlinks.ro/Forum/viewtopic.php?t=36018

Andrei wrote: Using Cauchy one can prove $(a+b+c)\left(\frac{(a-b)^2}{b}+\frac{(b-c)^2}{c}+\frac{(c-a)^2}{a} \right) \geq 4$ $($max$(a,b,c)-$min$(a,b,c)$$)^2$. q.e.d. I don't understand about it.Could you explain anymore?

$(a+b+c)\left(\frac{(a-b)^2}{b}+\frac{(b-c)^2}{c}+\frac{(c-a)^2}{a} \right) \geq (|a-b|+|b-c|+|c-a|)^2 $ WLOG $|c-a|=max(|a-b|,|b-c|,|c-a|)$ Then, We get $|a-b|+|b-c|\geq |c-a|$ So, $|a-b|+|b-c|+|c-a| \geq 2|c-a| $ Therefore, $(a+b+c)\left(\frac{(a-b)^2}{b}+\frac{(b-c)^2}{c}+\frac{(c-a)^2}{a} \right)\geq 4 max(|a-b|,|b-c|,|c-a|)^2 $.

Thank you very very much.

mathpk wrote: Thank you very very much. Equality hold iff one of twe cases occur : (1).a=b=c ; (2). c=ωb and a=ωc ,where ω=(5^1/2-1)/2.

With Lagrange you have a^2/b+b^2/c+ c^2/a - (a+b+c)= ( (ac-b^2)^2/bc + (bc - a^2)^2/ab + (ab-c^2)^2/ac )/a+b+c. So you have to prove that (ac-b^2)^2/bc + (bc - a^2)^2/ab + (ab-c^2)^2/ac >= 4(a-b)^2.But (ab-c^2)^2/ac >= 0 and (ac-b^2)^2/bc + (bc - a^2)^2/ab >= (ac-b^2-bc+a^2)^2/b(a+c) =(a-b)^2(a+b+c)^2/b(a+c). With AM-GM you have b(a+c) <= (a+b+c)^2/4 so (a-b)^2(a+b+c)^2/b(a+c) >= 4(a-b)^2. We get a^2/b+b^2/c+ c^2/a - (a+b+c) >= 4(a-b)^2/(a+b+c).

jivko777 wrote: Thanks zhaoli, but what I meant were the problems from BMO 2005 (you posted only the third one). Well anyways, they will certainly appear at some point. Regards, Zhivko you can find the questions of bmo 2005 at http://www.rms.unibuc.ro/balkan.html

ciprian wrote: With Lagrange you have a^2/b+b^2/c+ c^2/a - (a+b+c)= ( (ac-b^2)^2/bc + (bc - a^2)^2/ab + (ab-c^2)^2/ac )/a+b+c. who on earth can you obtain smth like this?

You just use Lagrange identity for 3 numbers

Andrei wrote: Rewrite the intial inequality to: $\frac{(a-b)^2}{b}+\frac{(b-c)^2}{c}+\frac{(c-a)^2}{a} \geq \frac{4(a-b)^2}{a+b+c}$. q.e.d. I guess the given inequality can be rewritten as : $\frac{(a^2-b^2)}{b}+\frac{(b^2-c^2)}{c}+\frac{(c^2-a^2)}{a} \geq \frac{4(a-b)^2}{a+b+c}$. Not as Andrei stated it to be.......Now how will you solve it?

Andrei's right. We do the calculation $a^2/b -2a+b$

I am so Dumb

Is this problem hard or easy for those who have solved it ?!?!?!

ciprian wrote: You just use Lagrange identity for 3 numbers i forgot about this! what's lagrange identity?

See attachment.

Attachments:

thanks i think i have seen it in Kedlaya's notes

ciprian wrote: With Lagrange you have a^2/b+b^2/c+ c^2/a - (a+b+c)= ( (ac-b^2)^2/bc + (bc - a^2)^2/ab + (ab-c^2)^2/ac )/a+b+c. So you have to prove that (ac-b^2)^2/bc + (bc - a^2)^2/ab + (ab-c^2)^2/ac >= 4(a-b)^2.But (ab-c^2)^2/ac >= 0 and (ac-b^2)^2/bc + (bc - a^2)^2/ab >= (ac-b^2-bc+a^2)^2/b(a+c) =(a-b)^2(a+b+c)^2/b(a+c). With AM-GM you have b(a+c) <= (a+b+c)^2/4 so (a-b)^2(a+b+c)^2/b(a+c) >= 4(a-b)^2. We get a^2/b+b^2/c+ c^2/a - (a+b+c) >= 4(a-b)^2/(a+b+c). Could someone explain it to me with more details?

ciprian wrote: With Lagrange you have$a^2/b+b^2/c+ c^2/a - (a+b+c)= ( (ac-b^2)^2/bc + \dfrac{(bc - a^2)^2}{ab} + \dfrac{(ab-c^2)^2}{ac} )-a+b+c$. So you have to prove that $(ac-b^2)^2/bc + (bc - a^2)^2/ab + (ab-c^2)^2/ac \geq 4(a-b)^2.But (ab-c^2)^2/ac \geq 0$ and $(ac-b^2)^2/bc + (bc - a^2)^2/ab \geq (ac-b^2-bc+a^2)^2/b(a+c) =(a-b)^2(a+b+c)^2/b(a+c)$. With AM-GM you have$b(a+c) \leq (a+b+c)^2/4 so (a-b)^2(a+b+c)^2/b(a+c) \geq 4(a-b)^2$. We get $a^2/b+b^2/c+ c^2/a - b+c) \geq 4(a-b)^2/(a+b+c)$.

We have $\dfrac{a^2}{b}+\dfrac{b^2}{c}+\dfrac{c^2}{a}\geq a+b+c+\dfrac{4(a-b)^2}{a+b+c} \Leftrightarrow$ $\dfrac{a^2}{b} + b - 2a +\dfrac{b^2}{c} + c - 2b +\dfrac{c^2}{a}+ a - 2c \geq \dfrac{4(a-b)^2}{a+b+c} \Leftrightarrow$ $\dfrac{(a-b)^2}{b} + \dfrac{(b-c)^2}{c} + \dfrac{(c-a)^2}{a} \geq \dfrac{4(a-b)^2}{a+b+c}$ Furthermore, by Cauchy, we have that $\dfrac{(a-b)^2}{b} + \dfrac{(a-b)^2}{a+c} \ge \dfrac{4(a-b)^2}{a+b+c}$ Therefore, its enough to prove that $\dfrac{(a-b)^2}{b} + \dfrac{(b-c)^2}{c} + \dfrac{(c-a)^2}{a} \geq \dfrac{(a-b)^2}{b} + \dfrac{(a-b)^2}{a+c} \Leftrightarrow$ $(a+c) \left ( \dfrac{(b-c)^2}{c} + \dfrac{(c-a)^2}{a} \right ) \ge (a-b)^2$, which is also Cauchy.

Let $a_1,a_2,\cdots,a_n$ be positive real numbers. Prove the inequality\[\frac{a_1^2}{a_2}+\frac{a_2^2}{a_3}+\cdots+ \frac{a_{n-1}^2}{a_n}+\frac{a_n^2}{a_1}\geq a_1+a_2+\cdots+a_n+\frac{4(a_1-a_2)^2}{a_1+a_2+\cdots+a_n}.\] $\Leftrightarrow $\[ \frac{(a_1-a_2)^2}{a_2}+\frac{(a_2-a_3)^2}{a_3}+\cdots+ \frac{(a_{n-1}-a_n)^2}{a_n}+\frac{(a_n-a_1)^2}{a_1}\geq \frac{4(a_1-a_2)^2}{a_1+a_2+\cdots+a_n}.\] http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&p=2995638

sqing wrote: Let $a_1,a_2,\cdots,a_n$ be positive real numbers. Prove the inequality\[\frac{a_1^2}{a_2}+\frac{a_2^2}{a_3}+\cdots+ \frac{a_{n-1}^2}{a_n}+\frac{a_n^2}{a_1}\geq a_1+a_2+\cdots+a_n+\frac{4(a_1-a_2)^2}{a_1+a_2+\cdots+a_n}.\] $\Leftrightarrow $\[ \frac{(a_1-a_2)^2}{a_2}+\frac{(a_2-a_3)^2}{a_3}+\cdots+ \frac{(a_{n-1}-a_n)^2}{a_n}+\frac{(a_n-a_1)^2}{a_1}\geq \frac{4(a_1-a_2)^2}{a_1+a_2+\cdots+a_n}.\] http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&p=2995638 Generalization Let $a_1,a_2,\cdots,a_n$ be positive real numbers and $r\in {{\mathbb{N}}^{*}}$. Prove the inequality \[\frac{a_{1}^{r+1}}{a_{2}^{r}}+\frac{a_{2}^{r+1}}{a_{3}^{r}}+\cdots +\frac{a_{n-1}^{r+1}}{a_{n}^{r}}+\frac{a_{n}^{r+1}}{a_{1}^{r}}\ge {{a}_{1}}+{{a}_{2}}+\cdots +{{a}_{n}}+\frac{4{{({{a}_{1}}-{{a}_{2}})}^{2}}}{{{a}_{1}}+{{a}_{2}}+\cdots +{{a}_{n}}}.\] http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&p=2995638

arqady wrote: $\sum_{cyc}\left(\frac{a^2}{b}-a\right)=\sum_{cyc}\left(\frac{a^2}{b}-2a+b\right)=\sum_{cyc}\frac{(a-b)^2}{b}$ $\geq\frac{(a-b+c-b+a-c)^2}{a+b+c}=\frac{4(a-b)^2}{a+b+c}$. http://www.artofproblemsolving.com/Forum/viewtopic.php?f=151&t=551876 http://www.artofproblemsolving.com/Forum/viewtopic.php?f=51&t=378359&p=2090931&hilit=inequality+2005#p2090931 http://www.artofproblemsolving.com/Forum/viewtopic.php?f=51&t=347475&hilit=inequality+2005 http://www.artofproblemsolving.com/Forum/viewtopic.php?f=51&t=377018

I learned too much andrei!thanks a lot:)

Fixing $ \text\LaTeX $ in ciprian's solution: By Lagrange we have $\frac{a^2}{b}+\frac{b^2}{c}+ \frac{c^2}{a} - (a+b+c)= \frac{ \frac{\left(ac-b^2 \right)^2}{bc} + \frac{\left(bc - a^2\right)^2}{ab} + \frac{\left(ab-c^2\right)^2}{ac}}{a+b+c}$ So we have to prove that $\frac{(ac-b^2)^2}{bc} + \frac{(bc - a^2)^2}{ab} + \frac{(ab-c^2)^2}{ac}\geq 4(a-b)^2$ But $\frac{(ab-c^2)^2}{ac}\geq 0$ and $\frac{(ac-b^2)^2}{bc} + \frac{(bc - a^2)^2}{ab} \geq \frac{(ac-b^2-bc+a^2)^2}{b(a+c)} = \frac{(a-b)^2(a+b+c)^2}{b(a+c)}$ By $\text{AM-GM}$ we have $b(a+c) \leq \frac{(a+b+c)^2}{4} \implies \frac{(a-b)^2(a+b+c)^2}{b(a+c)} \geq 4(a-b)^2$ Hence we get $\frac{a^2}{b}+\frac{b^2}{c}+ \frac{c^2}{a} - (a+b+c) \geq \frac{4(a-b)^2}{(a+b+c)}$

Note that we need to prove $(\frac{a^2}{b} - a) + (\frac{b^2}{c} - b) + (\frac{c^2}{a} - c) \ge \frac{4(a-b)^2}{a+b+c}$ or $\frac{a^2-ab}{b} + \frac{b^2-bc}{c} + \frac{c^2-ca}{a} \ge \frac{4(a-b)^2}{a+b+c}$ or $(\frac{a^2-ab}{b} + b - a) + (\frac{b^2-bc}{c} + c - b) + (\frac{c^2-ca}{a} + a - c) \ge \frac{4(a-b)^2}{a+b+c}$ or $\frac{(a-b)^2}{b} + \frac{(b-c)^2}{c} + \frac{(c-a)^2}{a} \ge \frac{4(a-b)^2}{a+b+c}$. Using titu we have $\frac{(a-b)^2}{b} + \frac{(b-c)^2}{c} + \frac{(c-a)^2}{a} \ge \frac{(|a-b)| + |b-c| + |c-a|)^2}{a+b+c} \ge \frac{4(a-b)^2}{a+b+c}$ so we need to prove $|a-b)| + |b-c| + |c-a| \ge 2(a-b)$ which is true.

Let $a,b,c$ be positive real numbers. Prove the inequality \[\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}>a+b-2c+\frac{5(a-b)^2}{a+b}.\]

Let $a,b,c$ be positive real numbers. Prove the inequality \[\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}>2(a+b-6c)+\frac{7(a-b)^2}{a+b}\]