shobber wrote: $x$, $y$ and $z$ are positive reals such that $x+y+z=xyz$. Find the minimum value of: \[ x^7(yz-1)+y^7(zx-1)+z^7(xy-1) \] Let $xy=\frac{1}{c},$ $xz=\frac{1}{b}$ and $yz=\frac{1}{a}.$ Then $x=\sqrt{\frac{a}{bc}},$ $y=\sqrt{\frac{b}{ac}}$ and $z=\sqrt{\frac{c}{ab}}.$ Hence,

shobber wrote: $x$, $y$ and $z$ are positive reals such that $x+y+z=xyz$. Find the minimum value of: \[ x^7(yz-1)+y^7(zx-1)+z^7(xy-1) \] Did you post to wrong place? 1. Obviously (10 second, Cauchy) 2. Chinese TST without solution??

langtupanda wrote: 2. Chinese TST without solution?? What's wrong with that?

Nobody solved it, so how it became the test? At least one who gave the test.

langtupanda wrote: Nobody solved it, so how it became the test? At least one who gave the test. I still do not know what you are talking about.

shobber wrote: $x$, $y$ and $z$ are positive reals such that $x+y+z=xyz$. Find the minimum value of: \[ x^7(yz-1)+y^7(zx-1)+z^7(xy-1) \] Let $xy=\frac{1}{c},$ $xz=\frac{1}{b}$ and $yz=\frac{1}{a}.$ Then $x=\sqrt{\frac{a}{bc}},$ $y=\sqrt{\frac{b}{ac}}$ and $z=\sqrt{\frac{c}{ab}}.$ Hence, $a+b+c=1$ and langtupanda wrote: Obviously (10 second, Cauchy)

langtupanda wrote: Nobody solved it, so how it became the test? At least one who gave the test. Quote: Problems you couldn't solve and to which you know that there is a solution (i.e. a problem from a contest, etc.) but you don't know it. shobber's just posting the problems here and probably he hasn't thought about it. So what's the problem with this?

langtupanda wrote: 2. Chinese TST without solution?? Why do you think that there is no solution, actually there is. But Shobber is just sending the problems, maybe he doesn't need the solutions. Maybe he is just with the problems from TST. . Davron Latipov

ahhh maybe you are confusing about sending this problem in the Unsolved section, there is no problem, if it is unsolved for the user he can send it as unsolved, since it is unsolved for him ... Davron Latipov

Sorry shobber I just mean you should post this problem to Solved Section. This is my solution: From x + y + z = x.y.z => x.y.z >= 3.[3^(1/2)] ( Cauchy) and 1/(xy) + 1/(yz) + 1/(zx) = 1 => 1/x.[(y+z)/(yz)] = [(yz - 1)/(yz)] => (yz - 1) = 1/x.(y + z) So we have the inequation similar to this one: x^6.(y + z) + y^6.(z + x) +z^6.(x + y) Cauchy 6 numbers and using x.y.z >= 3.[3^(1/2)] so we can have the minimum

langtupanda wrote: Sorry shobber I just mean you should post this problem to Unsolved Section. Never mind. I just like to post the Contest problems in the Unsolved section. From your solution I guess that you are from Vietnam since you call the inequality Cauchy but we call it AM-GM.

shobber wrote: langtupanda wrote: Sorry shobber I just mean you should post this problem to Unsolved Section. Never mind. I just like to post the Contest problems in the Unsolved section. From your solution I guess that you are from Vietnam since you call the inequality Cauchy but we call it AM-GM. Cool! you are right, but what is AM-GM?

AM-Gm is the inequality you called Cauchy These names ........ BTW , nice solution Arqady .

silouan wrote: AM-Gm is the inequality you called Cauchy These names ........ BTW , nice solution Arqady . I mean its real name, not its initials

langtupanda wrote: silouan wrote: AM-Gm is the inequality you called Cauchy These names ........ BTW , nice solution Arqady . I mean its real name, not its initials Arithmetic Mean-Geometric Mean Inequality, which states that the arithmetic mean of $n$ non-negative real numbers is greater or equal than the geometirc mean of those $n$ numbers. Equality holds when all the numbers are equal.

Does this work?

Same solution as AwesomeToad. This, for me, was probably one of the easiest China TST problems I have ever seen.

Alternatively, one could use Muirhead to solve this problem. As $x^7(yz-1) + y^7(zx-1)+z^7(xy-1) = x^6(xyz-x) + y^6(xyz-y) + z^6(xyz-z) = x^6(y+z) + y^6(z+x) + z^6(x+y),$ we have from the fact that $(6, 1, 0)$ majorizes $(3, 2, 2)$ and Muirhead that: $$x^7(yz-1) + y^7(zx-1) + z^7(xy-1) \ge 2x^2y^2z^2(x+y+z) = 2(x+y+z)^3. \qquad (1)$$ If we set $s = x+y+z,$ then we know that $s \ge 3 \cdot s^{\frac13} \Rightarrow s \ge 3 \sqrt 3.$ Hence $(1)$ yields that the sum in question is at least $162 \sqrt 3.$ Since this value is obtained at $x = y = z = \sqrt 3,$ we have shown that the answer is $\boxed{162 \sqrt 3}.$ $\square$

The original problem is a special case of the following generalization where $$p=2,\lambda =1$$

Generalization 1 Let $x,y,z$ be positive reals such that $x+y+z=\lambda xyz$. Then prove that $$\sum_{cyc}{\left(x^{3p+1}\left(\lambda yz-1\right)\right)}\geq \dfrac{2\left(3\sqrt{3}\right)^{p+1}}{\sqrt{\lambda ^{3p+1}}}$$

Generalization 2 Let $x_{1},x_{2},\cdots,x_{n},\lambda,p,n$ be positive reals ($n\geq 3$) such that $\sum_{cyc}{x_{1}}=\lambda \prod{x_{1}}$. Then prove that $$\sum_{cyc}{\left(x^{pn+1}\left(\lambda \dfrac{\prod{x_{i}}}{x_{1}}-1\right)\right)}\geq \dfrac{\left(n-1\right)\sqrt[n-1]{n^{n\left(p+1\right)}}}{\sqrt[n-1]{\lambda ^{pn+1}}}$$