very easy problem . CLear arithmetic, and AM-GM at the end

sandu2508 wrote: Let $a,b$ and $c$ be positive real numbers. Prove that \[ \frac{a^2b(b-c)}{a+b}+\frac{b^2c(c-a)}{b+c}+\frac{c^2a(a-b)}{c+a} \ge 0\] Substitution $x=\frac{1}{a}, y=\frac{1}{b}, z=\frac{1}{c},$ this inequality become $\sum\frac{x-z}{y+z}\ge0,$ which is true by Rearrangement Inequality $\sum x\left(\frac{1}{y+z}\right)\ge\sum z\left(\frac{1}{y+z}\right).$

An ugly solution: If somebody had the patience to make the common denominator, he would had a surprise: $\frac{a^{2}b(b-c)}{a+b}+\frac{b^{2}c(c-a)}{b+c}+\frac{c^{2}a(a-b)}{c+a} = \frac{a^3b^3-a^3b^2c+a^3c^3-a^2bc^3+b^3c^3-ab^3c^2}{(a+b)(b+c)(c+a)}.$ We have to prove that: $a^3b^3-a^3b^2c+a^3c^3-a^2bc^3+b^3c^3-ab^3c^2 \geq 0 \Leftrightarrow$ $a^3b^3+b^3c^3+c^3a^3 \geq a^3b^2c+a^2bc^3+ab^3c^2 $. We can see that this inequality is true by Muirhead, and prove it taking: $\frac{a^3b^3+a^3b^3+b^3c^3}{3} \geq a^2b^3c$ (by AM-GM). and its analogs. Summing up those 3 inequalities, our problem is solved.

Wow easy. Rearrangement is so obvious, the differences b-c,c-a,a-b immediately reminded me of it (wya, excellent solution, I never thought of that substitution). I thought like this: Dividing everything with abc gets rid of ugly $a^2b,b^2c,c^2a$ terms and leaves us only with $ab,bc,ca$ to work with. So, if some of $a,b,c$ is zero, easy. Else divide everything with abc, and set substitute $x=bc,y=ca,z=ab$ We get the inequality $\sum \frac{z-y}{x+y}\ge 0$ which is true by Rearrangement (wya's work)

very easy like this :if we sum up abc to each term of left hand side and also 3abc to the right hand side and then we apply the a.m-g.m...all is done

Let's do it, dear Anni! The inequality is equivalent to \[ abc+\frac{a^{2}b(b-c)}{a+b}+abc+\frac{b^{2}c(c-a)}{b+c}+abc+\frac{c^{2}a(a-b)}{c+a}\ge 3abc \] Therefore \[\left.LHS=ab^2\cdot\frac{a+c}{a+b}+bc^2\cdot\frac{b+a}{b+c}+ca^2\cdot\frac{c+b}{c+a}\stackrel{AM-GM}{\geq}3\sqrt[3]{ab^2\cdot\frac{a+c}{a+b}\cdot bc^2\cdot\frac{b+a}{b+c}\cdot ca^2\cdot\frac{c+b}{c+a}}=3abc=RHS\right.\]

What is BMO？ Balank or British

It's Balkan!

There are many ways to solve this! Another one is: Equivalently we have: $\sum_{cyclic}{\frac{ab}{bc + ca}} \geq \sum_{cyclic}{\frac{a}{b + a}}$ Add $\sum_{cyclic}{\frac{b}{b + a}}$ to both sides Then we need to show: $\sum_{cyclic}{\frac{ab}{ac + bc} + \frac{b}{b + a}} = \sum_{cyclic}{\frac{ab}{ac + bc} + \frac{bc}{bc + ac}} = \sum_{cyclic}{\frac{ab + bc}{ac + bc}} \geq 3$ which is true because by AM-GM: $\sum_{cyclic}{\frac{ab + bc}{ac + bc}} \geq 3\sqrt[3]{\frac{(ab + bc)(bc + ca)(ca + ab)}{(ab + ca)(ca + bc)(bc + ab)}} = 3$.

which state proposed this problem? was any problem in the exam proposed by albanian leader?

wya may I ask something? When you substitute the $x = 1/a$, $y = 1/b$, $z = 1/c$, did you have to write out, by hand, the transformation from the original substitution into this?: $\sum\frac{x-z}{y+z}\ge0$ Or is there some property of summations that allows you to make easy work of such a substitution? I did it by hand using the first term $\frac{a^{2}b(b - c)}{a + b}$ and got $\sum\frac{z-y}{y+x}\ge0$ which is equivalent to yours, however, I was wondering why you did not end up with the same looking summation as I had, and thus, was wondering if there was some special way of doing this. and can someone show me the rearrangement part?

The substitution that wya used is first that try most of the competitors leading us to the equivalent inequality $ \sum\frac{x-z}{y+z}\ge0 $ If we consider that the sum of the nominators is zero, we can continue even without using a theorem. Then we can re-write in obvious forms: if x is maximum(or one of the maximums) using $ (y-x)=-((z-y)+(x-z))$: \[ \frac{(z-y)^{2}}{(x+y)(x+z)}+\frac{(x-z)(x-y)}{(y+z)(x+z)}\ge 0 \] and it is clear that inequality holds. Using same idea, re-writing other two nominators, when y or z is maximum, we can obtain some similar forms. Thus, it holds in every case.

sandu2508 wrote: Let $a,b$ and $c$ be positive real numbers. Prove that \[ \frac{a^2b(b-c)}{a+b}+\frac{b^2c(c-a)}{b+c}+\frac{c^2a(a-b)}{c+a} \ge 0\] $\Leftrightarrow \frac{c^3(2a+b)(a-b)^2+a^3(2b+c)(b-c)^2+b^3(2c+a)(c-a)^2}{3}\ge 0$

turcas_c wrote: An ugly solution: If somebody had the patience to make the common denominator, he would had a surprise: $\frac{a^{2}b(b-c)}{a+b}+\frac{b^{2}c(c-a)}{b+c}+\frac{c^{2}a(a-b)}{c+a} = \frac{a^3b^3-a^3b^2c+a^3c^3-a^2bc^3+b^3c^3-ab^3c^2}{(a+b)(b+c)(c+a)}.$ We have to prove that: $a^3b^3-a^3b^2c+a^3c^3-a^2bc^3+b^3c^3-ab^3c^2 \geq 0 \Leftrightarrow$ $a^3b^3+b^3c^3+c^3a^3 \geq a^3b^2c+a^2bc^3+ab^3c^2 $. We can see that this inequality is true by Muirhead, and prove it taking: $\frac{a^3b^3+a^3b^3+b^3c^3}{3} \geq a^2b^3c$ (by AM-GM). and its analogs. Summing up those 3 inequalities, our problem is solved. I proved it just like you and I believe this is the easiest and most elegant way of proving this inequality.

Let $abc=1$, and then inequality becomes: \[ \sum_{cyc}{\frac{ab}{ac+bc}} \geq \sum_{cyc}{\frac{ac}{ac +bc}} \] Let $ab=x, bc=y, ca=z$, we have: \[ \frac{x}{y+z} +\frac{y}{z+x} +\frac{z}{x+y} \geq \frac{x}{x+y} +\frac{y}{y+z} +\frac{z}{z+x} \] Which is true by Rearrangement inequality. Edit: After reading what was written above, I realized that my solution is pretty much what Bugi said.

turcas_c wrote: An ugly solution: If somebody had the patience to make the common denominator, he would had a surprise: $\frac{a^{2}b(b-c)}{a+b}+\frac{b^{2}c(c-a)}{b+c}+\frac{c^{2}a(a-b)}{c+a} = \frac{a^3b^3-a^3b^2c+a^3c^3-a^2bc^3+b^3c^3-ab^3c^2}{(a+b)(b+c)(c+a)}.$ We have to prove that: $a^3b^3-a^3b^2c+a^3c^3-a^2bc^3+b^3c^3-ab^3c^2 \geq 0 \Leftrightarrow$ $a^3b^3+b^3c^3+c^3a^3 \geq a^3b^2c+a^2bc^3+ab^3c^2 $. We can see that this inequality is true by Muirhead, and prove it taking: $\frac{a^3b^3+a^3b^3+b^3c^3}{3} \geq a^2b^3c$ (by AM-GM). and its analogs. Summing up those 3 inequalities, our problem is solved. My first try ... is same as your solution. Nice AM-GM in the end!

turcas_c wrote: An ugly solution: If somebody had the patience to make the common denominator, he would had a surprise: $\frac{a^{2}b(b-c)}{a+b}+\frac{b^{2}c(c-a)}{b+c}+\frac{c^{2}a(a-b)}{c+a} = \frac{a^3b^3-a^3b^2c+a^3c^3-a^2bc^3+b^3c^3-ab^3c^2}{(a+b)(b+c)(c+a)}.$ We have to prove that: $a^3b^3-a^3b^2c+a^3c^3-a^2bc^3+b^3c^3-ab^3c^2 \geq 0 \Leftrightarrow$ $a^3b^3+b^3c^3+c^3a^3 \geq a^3b^2c+a^2bc^3+ab^3c^2 $. We can see that this inequality is true by Muirhead, and prove it taking: $\frac{a^3b^3+a^3b^3+b^3c^3}{3} \geq a^2b^3c$ (by AM-GM). and its analogs. Summing up those 3 inequalities, our problem is solved. i did it tis way and i took 1 point:S

Ledio9494 wrote: turcas_c wrote: An ugly solution: If somebody had the patience to make the common denominator, he would had a surprise: $\frac{a^{2}b(b-c)}{a+b}+\frac{b^{2}c(c-a)}{b+c}+\frac{c^{2}a(a-b)}{c+a} = \frac{a^3b^3-a^3b^2c+a^3c^3-a^2bc^3+b^3c^3-ab^3c^2}{(a+b)(b+c)(c+a)}.$ We have to prove that: $a^3b^3-a^3b^2c+a^3c^3-a^2bc^3+b^3c^3-ab^3c^2 \geq 0 \Leftrightarrow$ $a^3b^3+b^3c^3+c^3a^3 \geq a^3b^2c+a^2bc^3+ab^3c^2 $. We can see that this inequality is true by Muirhead, and prove it taking: $\frac{a^3b^3+a^3b^3+b^3c^3}{3} \geq a^2b^3c$ (by AM-GM). and its analogs. Summing up those 3 inequalities, our problem is solved. i did it tis way and i took 1 point:S If you did it like this than you must have taken 10 points! What did your leader say?

Ledio9494 wrote: turcas_c wrote: An ugly solution: If somebody had the patience to make the common denominator, he would had a surprise: $\frac{a^{2}b(b-c)}{a+b}+\frac{b^{2}c(c-a)}{b+c}+\frac{c^{2}a(a-b)}{c+a} = \frac{a^3b^3-a^3b^2c+a^3c^3-a^2bc^3+b^3c^3-ab^3c^2}{(a+b)(b+c)(c+a)}.$ We have to prove that: $a^3b^3-a^3b^2c+a^3c^3-a^2bc^3+b^3c^3-ab^3c^2 \geq 0 \Leftrightarrow$ $a^3b^3+b^3c^3+c^3a^3 \geq a^3b^2c+a^2bc^3+ab^3c^2 $. We can see that this inequality is true by Muirhead, and prove it taking: $\frac{a^3b^3+a^3b^3+b^3c^3}{3} \geq a^2b^3c$ (by AM-GM). and its analogs. Summing up those 3 inequalities, our problem is solved. i did it tis way and i took 1 point:S If you used Muirhead, it would be wrong, Muirhead guarantees inequalities for symmetric sums but not cyclic. So Muirhead only proves that $a^3b^3+a^3c^3+b^3a^3+b^3c^3+c^3a^3+c^3b^3 \ge a^3b^2c+a^3c^2b+b^3a^2c+b^3c^2a+c^3a^2b+c^3b^2a$ If you used AM-GM, it would be wrong, AM-GM proves that: $\frac{a^3b^3+a^3b^3+b^3c^3}{3} \ge b^3a^2c$ $\frac{b^3c^3+b^3c^3+c^3a^3}{3} \ge c^3b^2a$ $\frac{c^3a^3+c^3a^3+a^3b^3}{3} \ge a^3c^2b$ So $a^3b^3+b^3c^3+c^3a^3 \ge a^3c^2b+b^3a^2c+c^3b^2a$ But you were trying to prove that $a^3b^3+b^3c^3+c^3a^3 \ge a^3b^2c+b^3c^2a+c^3a^2b$ Which is different from the above.

Duarti wrote: By C-S: $\sum{\frac{a^2b^2(b-c)^2}{b(a+b)(b-c)}}\geq \frac{(\sum{ab(b-c)})^2}{\sum{b(a+b)(b-c)}}$ It's wrong of course.

With the rearrangement inequality, we know that $$x^3+y^3+z^3\ge x^2y+y^2z+z^2x$$ for all positive real $x, y, z$. Now we let $a=\frac{1}{x}, b=\frac{1}{y}, c=\frac{1}{z}$ so we know that $$\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}\ge\frac{1}{bc^2}+\frac{1}{ca^2}+\frac{1}{ab^2}$$ for all positive real $a, b, c$. Multiplying both sides by $a^3b^3c^3$ we get $$a^3b^3+b^3c^3+c^3a^3\ge a^3b^2c+b^3c^2a+c^3a^2b.$$ Therefore $a^3b^3+b^3c^3+c^3a^3-a^3b^2c-b^3c^2a-c^3a^2b\ge0$. Adding some stuff and taking it away again, we get $$a^2b^3c-a^2bc^3+a^3b^3-a^3bc^2+ab^2c^3+b^3c^3-a^3b^2c-a^2b^3c+a^3bc^2-ab^3c^2+a^3c^3-ab^2c^3 \ge 0.$$ Oh, look! It magically factorises to $$a^2b(b^2-c^2)(c+a)+b^2(c^2-a^2)(a+b)+c^2a(a^2-b^2)(b+c)\ge0.$$ I bet you're wondering how I found such a beautiful factorisation. Well, now, you can divide both sides by $(a+b)(b+c)(c+a)$ to get \[ \frac{a^2b(b-c)}{a+b}+\frac{b^2c(c-a)}{b+c}+\frac{c^2a(a-b)}{c+a} \ge 0. \] To find how I worked this (pretty boring) method out, just follow all my steps in reverse.

sandu2508 wrote: Let $a,b$ and $c$ be positive real numbers. Prove that \[ \frac{a^2b(b-c)}{a+b}+\frac{b^2c(c-a)}{b+c}+\frac{c^2a(a-b)}{c+a} \ge 0. \] Let $a,b,c>0$. Prove that\[\frac{a(c-b)}{2a+3b+c}+\frac{b(a-c)}{2b+3c+a}+\frac{c(b-a)}{2c+3a+b}\ge 0\]https://artofproblemsolving.com/community/c6h489134p2741648 \[\frac{(a+2b+1)(a-b)}{b(2c+a)}+\frac{(b+2c+1)(b-c)}{c(2a+b)}+\frac{(c+2a+1)(c-a)}{a(2b+c)}\ge0\]https://artofproblemsolving.com/community/c6h576961p3401520 \[\frac{a(a-b)}{c^2+ca+ab}+\frac{b(b-c)}{a^2+ab+bc}+\frac{c(c-a)}{b^2+bc+ca}\geq 0\]https://artofproblemsolving.com/community/c6h583859p3451999 For $a, b, c>0$, prove the following "twin"" inequalities: \[ \frac{ab(b^2-ac)}{a+b}+\frac{bc(c^2-ab)}{b+c}+\frac{ca(a^2-bc)}{c+a} \ge 0. \]\[ \frac{ab(ab-c^2)}{b+c}+\frac{bc(bc-a^2)}{c+a}+\frac{ca(ca-b^2)}{a+b} \ge 0. \]https://artofproblemsolving.com/community/c6h2094743p15141068

sandu2508 wrote: Let $a,b$ and $c$ be positive real numbers. Prove that \[ \frac{a^2b(b-c)}{a+b}+\frac{b^2c(c-a)}{b+c}+\frac{c^2a(a-b)}{c+a} \ge 0. \] By clearing the denominators, we have to show that \[\sum_{cyc} a^2b(b^2-c^2)(c+a)=\sum_{cyc} a^3b^3-\sum_{cyc} a^3b^2c\ge 0\]which by Schur's Inequality, we know that \[\sum_{cyc} a^3b^3 \ge \sum_{sym} a^3b^2c - 3a^2b^2c^2\ge \sum_{cyc} a^3b^2c\]$\textbf{\textit{IFF}}$ \[\sum_{cyc} a^2c \ge 3abc\]which is just AM-GM. $\quad \blacksquare$

By adding $abc$ to each fraction, we wish to prove $\sum \frac{ab^2(a+c)}{a+b} \geq 3abc$ which follows directly from AM-GM.

Let $a,b$ and $c$ be positive real numbers. Prove that$$ \frac{a^2b(ab-c^2)}{a+b}+\frac{b^2c(bc-a^2)}{b+c}+\frac{c^2a(ca-b^2)}{c+a} \ge 0$$

sqing wrote: Let $a,b$ and $c$ be positive real numbers. Prove that$$ \frac{a^2b(ab-c^2)}{a+b}+\frac{b^2c(bc-a^2)}{b+c}+\frac{c^2a(ca-b^2)}{c+a} \ge 0$$ It's $$\sum_{cyc}(a^4b^3-a^3b^2c^2)\geq0,$$which is obvious.

sandu2508 wrote: Let $a,b$ and $c$ be positive real numbers. Prove that \[ \frac{a^2b(b-c)}{a+b}+\frac{b^2c(c-a)}{b+c}+\frac{c^2a(a-b)}{c+a} \ge 0. \] if we divide each side by abc, we get $\sum(\frac{ab-ac}{ac+bc})\ge 0$ $ab=x, bc=y, ac=z$ $\sum(\frac{x-z}{y+z})\ge 0$ add 3 both sides, $1-\frac{z}{y+z}=\frac{y}{y+z}$ so, $\sum(\frac{x+y}{y+z})\ge 3$ and it is well-known(according to AM-GM). So, we are done.

Generalization 1 Let $a_{1},a_{2},\cdots,a_{n}$ be positive reals ($n\geq 3$). Then prove that $$\sum_{cyc}{\left(\dfrac{\prod{\left(a_{1}\right)}\dfrac{a_{k}}{a_{k-1}}\left(a_{k+1}-a_{k-1}\right)}{a_{k}+a_{k+1}}\right)}\geq 0$$

First sum it and after that it can be done easily by muirhead ineq

MexicOMM wrote: Sorry to revive, but I think this solution is the simplest one. Like everybody has seen, we need to prove that: \[ \frac{a^2b^2}{a+b} + \frac{a^2c^2}{a+c} + \frac{b^2c^2}{b+c} \ge \frac{a^2bc}{a+b} +\frac{abc^2}{a+c}+\frac{ab^2c}{b+c}\]WLOG we can asume that $a \ge b \ge c$, so we get $ \frac{ab}{a+b} \ge \frac{ac}{a+c} \ge \frac{cb}{c+b}$, you can see this by just simplifying, and $ab \ge ac \ge bc$, so applying the rearrangement inequality, with the previous sets we get the result. Why we can take $a\ge b\ge c$? This inequality not symmetric

BreezeCrowd wrote: MexicOMM wrote: Sorry to revive, but I think this solution is the simplest one. Like everybody has seen, we need to prove that: \[ \frac{a^2b^2}{a+b} + \frac{a^2c^2}{a+c} + \frac{b^2c^2}{b+c} \ge \frac{a^2bc}{a+b} +\frac{abc^2}{a+c}+\frac{ab^2c}{b+c}\]WLOG we can asume that $a \ge b \ge c$, so we get $ \frac{ab}{a+b} \ge \frac{ac}{a+c} \ge \frac{cb}{c+b}$, you can see this by just simplifying, and $ab \ge ac \ge bc$, so applying the rearrangement inequality, with the previous sets we get the result. Why we can take $a\ge b\ge c$? This inequality not symmetric It means triples $(ab,ac,bc)$ and $\left(\frac{ab}{a+b},\frac{ac}{a+c},\frac{cb}{c+b}\right)$ have the same ordering.