I remember Manlio asked me for the solution and I sent it to him I remember this problem because of nice equality case: $a=\frac{y+z}{2}$, $y=\frac{x+z}{2}$ and $c=\frac{x+y}{2}$

Case min{a,b,c}=0 is trivial. Assume that there are exist x,y,z,a,b,c satisfying min {a,b.c,}>0 and ax^2+by^2+cz^2+xyz< 4abc.We will prove that x+y+z < a+b+c(1). First,we prove that:If ax^2+by^2+cz^2+xyz=4abc then x+y+z\leq\ a+b+c ax^2+by^2+cz^2+xyz=4abc<=>\frac{x^2}{4bc}+\frac{y^2}{4ca}+\frac{z^2}{4xy}+\frac{xyz}{4abc}=1.Then,exist a triangle ABC satisfying:cosA=\sqrt \frac{x^2}{4bc}...(because we have the equality:(cosA)^2+(cosB)^2+(cosC)^2+2cosAcosBcosC=1)And x+y+z\leq\ a+b+c is equivalent to 2(\sqrt {bc}cosA+\sqrt{ca}cosB+\sqrt{ab}cosC\leq\ a+b+c,which is trivial. Now,from:ax^2+by^2+cz^2+xyz<4abc,put ax^2+by^2+cz^2=M;xyz=N.F(t)=Mt^2+Nt^3(t>0).F(t) is increasing function,F(1)<4abc.Then,there are exist 1<t so that F(t)=4abc.Now we can replace (x,y,z) by (tx,ty,tz) and see that x+y+z<t(x+y+z)\leq\ a+b+c.And we are done.

To me it's extremely beautiful!

Oh, I looked too fast at the solution. But I remembered by solution: suppose that the inequality of Vasc is not true and put $x=2u\sqrt{bc}, y=2v\sqrt{ca}; z=2w\sqrt{ab}$. Then we have $u^2+v^2+w^2+2uvw<1$ and $ a+b+c=2u\sqrt{bc}+2v\sqrt{ca}+2w\sqrt{ab}$. From the first relation there exists a triangle ABC and r smaller than 1 such that $ u=rcosA$ and so on. You will contradict then the famous inequality $A^2+B^2+C^2\geq 2AB cos T+2BC cos S+2CA cos Q$ which is true in any triangle TSQ and any positive A,B,C.

And as always I used a general method which immediately gives that extremal point $(a,b,c,x,y,z)$ satisfies the relation mentioned in my previous reply, hence the result.

xtar wrote: Find all of the positive real numbers like $ x,y,z,$ such that : 1.) $ x + y + z = a + b + c$ 2.) $ 4xyz = a^2x + b^2y + c^2z + abc$ Proposed to Gazeta Matematica in the 80s by VASILE CÎRTOAJE and then by Titu Andreescu to IMO 1995. $ a + b + c = x + y + z\Longrightarrow a^2x + b^2y + c^2z + abc - 4xyz$ $ = \frac {(a^2x + b^2y + c^2z)(x + y + z)^2}{(a + b + c)^2} + \frac {abc(x + y + z)^3}{(a + b + c)^3} - 4xyz$ $ = \sum{\left[\frac {x}{(a + b + c)^2} + \frac {bc(x + y + z)}{(a + b + c)^4}\right]\Big[y(a + b) - z(c + a)\Big]^2}$ $ + \frac {x + y + z}{2(a + b + c)^4}\sum{\Big[xa(b - c) - (yb - zc)(b + c)\Big]^2}\geq 0,$ with equality if and only if $ a = y + z - x,b = z + x - y,c = x + y - z.$ See here: http://www.mathlinks.ro/viewtopic.php?t=291232

ji chen how do you make such incredible sum of squares?

xtar wrote: Find all of the positive real numbers like $ x,y,z,$ such that : 1.) $ x + y + z = a + b + c$ 2.) $ 4xyz = a^2x + b^2y + c^2z + abc$ Proposed to Gazeta Matematica in the 80s by VASILE CÎRTOAJE and then by Titu Andreescu to IMO 1995. If someone want to see a nice proof by Cauchy-Schwarz for this problem, have a look at Problem 4.20 - page 206 in new book: Vasile Cirtoaje, Vo Quoc Ba Can, Tran Quoc Anh, Inequalities with Beautitful Solutions, GIL publishing house 2009.

I saw a solution with lagrange. There you get the system $4yz-a^2=4xz-b^2=4xy-c^2$ and $x+y+z=a+b+c$. In the solution they just say that you'll get the result after solving the system, my questions is how to solve this system.

Glad you asked, bearkiller!

can_hang2007 wrote: xtar wrote: Find all of the positive real numbers like $ x,y,z,$ such that : 1.) $ x + y + z = a + b + c$ 2.) $ 4xyz = a^2x + b^2y + c^2z + abc$ Proposed to Gazeta Matematica in the 80s by VASILE CÎRTOAJE and then by Titu Andreescu to IMO 1995. If someone want to see a nice proof by Cauchy-Schwarz for this problem, have a look at Problem 4.20 - page 206 in new book: Vasile Cirtoaje, Vo Quoc Ba Can, Tran Quoc Anh, Inequalities with Beautitful Solutions, GIL publishing house 2009. dear can : this web page cannot be open now could u please tell me where i can download this book ???

I'll prove the following statement instead: Modified statement wrote: Let $a,b,c,x,y,z$ be positive reals such that $a+b+c=x+y+z$. Prove that \[a^2x+b^2y+c^2z+abc\ge 4xyz\]with equality if and only if $(x,y,z)=\left(\frac{b+c}{2},\frac{a+c}{2},\frac{a+b}{2}\right)$. Homogenize the inequality to get \[\left[\left(a^2x+b^2y+c^2z\right)\left(a+b+c\right)+abc\left(x+y+z\right)\right]\left(x+y+z\right)^2\ge 4xyz\left(a+b+c\right)^3\]Note that \begin{align*} &\left(a^2x+b^2y+c^2z\right)\left(a+b+c\right)+abc\left(x+y+z\right)\\ &=\sum x\left[a^2\left(a+b+c\right)+abc\right]\\ &=\sum ax\left(a+b\right)\left(a+c\right) \end{align*} Hence the inequality is equivalent to \begin{align*} &\left[\sum ax\left(a+b\right)\left(a+c\right)\right]\left(\sum x\right)^2\ge 4xyz\left(a+b+c\right)^3\\ &\iff\left[\sum\left(\frac{a}{a+b+c}\cdot\frac{x}{b+c}\right)\right]\left[\sum\left(\frac{b+c}{2\left(a+b+c\right)}\cdot\frac{x}{b+c}\right)\right]^2\ge\prod\frac{x}{b+c} \end{align*} Let $(p,q,r)=\left(\frac{x}{b+c},\frac{y}{a+c},\frac{z}{a+b}\right)$. Then the inequality is equivalent to \[\left[\sum\left(\frac{a}{a+b+c}\cdot p\right)\right]\left[\sum\left(\frac{b+c}{2\left(a+b+c\right)}\cdot p\right)\right]^2\ge\prod p\] By weighted AM-GM inequality we have \begin{align*} &\left[\sum\left(\frac{a}{a+b+c}\cdot p\right)\right]\left[\sum\left(\frac{b+c}{2\left(a+b+c\right)}\cdot p\right)\right]^2\\ &\ge\left[\prod p^{\left(\frac{a}{a+b+c}\right)}\right]\left[\prod p^{\left(\frac{b+c}{2\left(a+b+c\right)}\right)}\right]^2\\ &=\left[\prod p^{\left(\frac{a}{a+b+c}\right)}\right]\left[\prod p^{\left(\frac{b+c}{a+b+c}\right)}\right]\\ &=\prod p \end{align*} Equality holds if and only if $p=q=r$, or equivalently $\frac{x}{b+c}=\frac{y}{a+c}=\frac{z}{a+b}$. Since $a+b+c=x+y+z$, this forces $\frac{x}{b+c}=\frac{y}{a+c}=\frac{z}{a+b}=\frac{1}{2}$, or equivalently $(x,y,z)=\left(\frac{b+c}{2},\frac{a+c}{2},\frac{a+b}{2}\right)$.

Very nice proof.