$x\neq y \neq z \neq x ?$.

rmtf1111 wrote: $x\neq y \neq z \neq x ?$. He mean all of them are different. sqing wrote: Let $x,y,z$ be positive integers such that $x\neq y\neq z$ .Prove that $$(x+y+z)(xy+yz+zx-2)\geq 9xyz.$$When does the equality hold? JBMO 2017, Q2 WLOG let $x<y<z$,$y=x+a,z=x+a+b$, obviously $a,b\ge 1$ and the ineq become $2(a^2+ab+b^2-3)x+(2a+b)(a^2+ab-2)\ge 0$ which is obviously true because $a,b\ge 1$ The equality holds when $x,y,z$ are consecutive positive integers. Edit @below,thanks!

This problem was proposed by me.

dangerousliri wrote: This problem was proposed by me. Thanks.

sqing wrote: Let $x,y,z$ be positive integers such that $x\neq y\neq z$ .Prove that $$(x+y+z)(xy+yz+zx-2)\geq 9xyz.$$When does the equality hold? JBMO 2017, Q2 $$(x+y+z)(xy+yz+zx-2)\geq 9xyz\iff x(y-z)^2+y(x-z)^2+z(x-y)^2\geq 2(x+y+z) $$

TLP.39 wrote: rmtf1111 wrote: $x\neq y \neq z \neq x ?$. He mean all of them are different. sqing wrote: Let $x,y,z$ be positive integers such that $x\neq y\neq z$ .Prove that $$(x+y+z)(xy+yz+zx-2)\geq 9xyz.$$When does the equality hold? JBMO 2017, Q2 WLOG let $x<y<z$,$y=x+a,z=x+a+b$, obviously $a,b\ge 1$ and the ineq become $2(a^2+ab+b^2-3)x+b(a^2+ab-2)\ge 0$ which is obviously true because $a,b\ge 1$ The equality holds when $x,y,z$ are consecutive positive integers. I think you did a little mistake should be $2(a^2+ab+b^2-3)x+(2a+b)(a^2+ab-2)\geq 0$ but still is true.

dangerousliri wrote: This problem was proposed by me. Congratulations! This problem is solved by me ^^

I think that this part will be difficult for juniors. $(2a+b)(a^2+ab-2)$

bitrak wrote: I think that this part will be difficult for juniors. $(2a+b)(a^2+ab-2)$ That is equivalent to $2a^3+3a^2b+ab^2-4a-2b$ so even if they can't factorise they can just finish from $a\geq 1$ and $b\geq 1$.

Yes. I agree. Many juniors said that problem was difficult. It is because they like to solve classical inequalities with Cauchy, Schur etc. and don't think out of frames. Your problem is very nice. Something similar (inequalities in N, Z, Q) I had posted here on 20 Oct 2016 https://artofproblemsolving.com/community/q1h1329045p7155572

It's a nice problem! Congratulations to dangerousliri! The problem reminds me the following: https://artofproblemsolving.com/community/c6h1245207p6383119

silouan wrote: It's a nice problem! Congratulations to dangerousliri! The problem reminds me the following: https://artofproblemsolving.com/community/c6h1245207p6383119 I actually got inspired from another inequality with the distinct integers but I couldn't find it.

sqing wrote: Let $x,y,z$ be positive integers such that $x\neq y\neq z \neq x$ .Prove that $$(x+y+z)(xy+yz+zx-2)\geq 9xyz.$$When does the equality hold? Proposed by Dorlir Ahmeti, Albania The original inequality is equivalent to $$(x+y+z)(xy+yz+zx) -9xyz\ge 2(x+y+z),$$or $$x(y-z)^2+y(x-z)^2+z(x-y)^2 \ge 2(x+y+z).$$Note that among $x+y-2z,\, y+z-2x,\, z+x-2y,$ there are two numbers having the same sign. Without loss of generality, assume that $(x+y-2z)(x+z-2y) \ge 0.$ We have $$2(x+y+z) \le 2(y-z)^2(x+y+z).$$Therefore, it suffices to prove that $$y(x-z)^2+z(x-y)^2 \ge x(y-z)^2+2(y+z)(y-z)^2,$$or $$(y+z)(x-y)(x-z) \ge 2(y+z)(y-z)^2.$$The last inequality is equivalent to $$(x-y)(x-z) -2(y-z)^2 \ge 0,$$or $$(x+z-2y)(x+y-2z) \ge 0,$$which is true.

It's too easy problem,also nice! WLOG,we assume that $x>y>z.$ Let's say $x=a+2,y=b+1,z=c.$ $\Longrightarrow$ $a \geq b \geq c.$ Then \[(x+y+z)(xy+yz+zx-2)\geq 9xyz\]$\Longleftrightarrow$ \[\sum_{cyc}{ab(a+b)}+a^2+2b^2+3c^2+3(a+2b+3c)+6ab\geq 10bc+2ac+18c+6abc\]$\Longleftrightarrow$ \[\sum_{cyc}{ab(a+b)}+a^2+2b^2+3c^2+3a+6b+6ab\geq 10bc+2ac+9c+6abc\]Clearly,since $LHS\geq 6abc+a^2+3c^2+6bc+3c+6b+2b^2$ So,it remains to prove that \[a^2+3c^2+6bc+3c+6b+2b^2\geq 10bc+2ac+9c\]$\Longleftrightarrow$ \[a^2+c^2+2c^2+6b+2b^2\geq 2ac+4bc+6c \]We only need to prove that $2b^2+2c^2+6b\geq 4bc+6c,$ which is obvious,so we're done.Equality holds iff $a=b=c.$

sqing wrote: Let $x,y,z$ be positive integers such that $x\neq y\neq z \neq x$ .Prove that $$(x+y+z)(xy+yz+zx-2)\geq 9xyz.$$When does the equality hold? Proposed by Dorlir Ahmeti, Albania here

sqing wrote: Let $x,y,z$ be positive integers such that $x\neq y\neq z \neq x$ .Prove that $$(x+y+z)(xy+yz+zx-2)\geq 9xyz.$$When does the equality hold? Proposed by Dorlir Ahmeti, Albania Nice problem !! and congratulations Dorlir Ahmeti

sqing wrote: Let $x,y,z$ be positive integers such that $x\neq y\neq z \neq x$ .Prove that $$(x+y+z)(xy+yz+zx-2)\geq 9xyz.$$When does the equality hold? Proposed by Dorlir Ahmeti, Albania

sqing wrote: Let $x,y,z$ be positive integers such that $x\neq y\neq z \neq x$ .Prove that $$(x+y+z)(xy+yz+zx-2)\geq 9xyz.$$When does the equality hold? Proposed by Dorlir Ahmeti, Albania Another solution: We can suppose $ x\geq y+1\geq z+2$ our inequality transform to $(z(x-y)+y(x-z)+3y-2)(x-y-1)+(x(y-z)+y(x-z)+3y+2)(y-z-1)\geq 0$ Which is obviusly true.

dangerousliri wrote: sqing wrote: Let $x,y,z$ be positive integers such that $x\neq y\neq z \neq x$ .Prove that $$(x+y+z)(xy+yz+zx-2)\geq 9xyz.$$When does the equality hold? Proposed by Dorlir Ahmeti, Albania Another solution: We can suppose $ x\geq y+1\geq z+2$ our inequality transform to $(z(x-y)+y(x-z)+3y-2)(x-y-1)+(x(y-z)+y(x-z)+3y+2)(y-z-1)\geq 0$ Which is obviusly true. Sir, was this the official solution ?

adityaguharoy wrote: dangerousliri wrote: sqing wrote: Let $x,y,z$ be positive integers such that $x\neq y\neq z \neq x$ .Prove that $$(x+y+z)(xy+yz+zx-2)\geq 9xyz.$$When does the equality hold? Proposed by Dorlir Ahmeti, Albania Another solution: We can suppose $ x\geq y+1\geq z+2$ our inequality transform to $(z(x-y)+y(x-z)+3y-2)(x-y-1)+(x(y-z)+y(x-z)+3y+2)(y-z-1)\geq 0$ Which is obviusly true. Sir, was this the official solution ? Nope. The official solution was that is posted by sqing here in 18-th post.

I tried inducting it at the contest and got 7 points. Heard that the jury laughed when they saw it.

dangerousliri wrote: This problem was proposed by me. Will you propose a problem this IMO ?)

rightways wrote: dangerousliri wrote: This problem was proposed by me. Will you propose a problem this IMO ?) You should wait another 20 days and I will tell you.

Let $x,y,z$ be positive integers such that $x\neq y\neq z \neq x$ .Prove that $$xyz(x^2+y^2+z^2-2) \ge 12(2x+2y+2z-3).$$When does the equality hold? (Datis-Kalali)

sqing wrote: Let $x,y,z$ be positive integers such that $x\neq y\neq z \neq x$ .Prove that $$xyz(x^2+y^2+z^2-2) \ge 12(2x+2y+2z-1).$$When does the equality hold? (Datis-Kalali) Inequality is not correct for x=1, y=2, z=3

bitrak wrote: sqing wrote: Let $x,y,z$ be positive integers such that $x\neq y\neq z \neq x$ .Prove that $$xyz(x^2+y^2+z^2-2) \ge 12(2x+2y+2z-1).$$When does the equality hold? (Datis-Kalali) Inequality is not correct for x=1, y=2, z=3 Unfortunate. But can we find the truth condition ?

sqing wrote: Let $x,y,z$ be positive integers such that $x\neq y\neq z \neq x$ .Prove that $$xyz(x^2+y^2+z^2-2) \ge 12(2x+2y+2z-3).$$When does the equality hold? (Datis-Kalali) It has been modified.

sqing wrote: Let $x,y,z$ be positive integers such that $x\neq y\neq z \neq x$ .Prove that $$xyz(x^2+y^2+z^2-2) \ge 12(2x+2y+2z-3).$$When does the equality hold? (Datis-Kalali) The main idea is similar to the one of the original problem. Wlog assume $z> y > x$, then $z\geq3$, $x\geq 2$, $x\geq 1$ and so $xyz\geq 6$. So, $$ xyz(x^2+y^2+z^2-2) \geq 6(x^2+y^2+z^2-2)$$Hence it is enough to show that $$6(x^2+y^2+z^2-2) \geq 24(x+y+z-3) \Leftrightarrow (x-2)^2+(y-2)^2+(z-2)^2 \geq 2$$which is obviously holds beacause $x\neq y\neq z \neq x$. Equality at $x=1,y=2,z=3$.

Solution $x\ge y\ge z$, The problem reduces to $\sum_{\text{cyc}}$ $y^2(x^2–z^2)^2 \ge $ $\sum_{\text{cyc}}$ ${2x}$ A true bcz 1) proof $y^2> 2$, $(x^2–z^2)^2\ge x$ since $x\ge 0$

sqing wrote: Let $x,y,z$ be positive integers such that $x\neq y\neq z \neq x$ .Prove that $$(x+y+z)(xy+yz+zx-2)\geq 9xyz.$$When does the equality hold? Proposed by Dorlir Ahmeti, Albania oneline down! let a>b>c,then a-b>=1,b-c>=1,a-c>=2, $\left( b+c \right)\left( \left( a-b \right)\left( a-c \right)-2 \right)+2a\left( {{\left( b-c \right)}^{2}}-1 \right)\ge 0$

sqing wrote: sqing wrote: Let $x,y,z$ be positive integers such that $x\neq y\neq z$ .Prove that $$(x+y+z)(xy+yz+zx-2)\geq 9xyz.$$When does the equality hold? JBMO 2017, Q2 $$(x+y+z)(xy+yz+zx-2)\geq 9xyz\iff x(y-z)^2+y(x-z)^2+z(x-y)^2\geq 2(x+y+z) $$ how do you know this x(y-z)^2+y(x-z)^2+z(x-y)^2 is bigger than or equel 2(x+y+z)?

He just multiplied the first inequality out, and rearranged it into the second inequality , he did not prove anything with that..

sqing wrote: Let $x,y,z$ be positive integers such that $x\neq y\neq z \neq x$ .Prove that $$(x+y+z)(xy+yz+zx-2)\geq 9xyz.$$When does the equality hold? Proposed by Dorlir Ahmeti, Albania Since the equation is symmetric and $x,y,z$ are distinct integers WLOG we can assume that $x\geq y+1\geq z+2$. \begin{align*} x+y+z\geq 3(z+1)\\ xy+yz+xz-2 = y(x+z)+xy-2 \geq (z+1)(2z+z)+z(z+2)-2 \\ xy+yz+xz-2 \geq 3z(z+2) \end{align*}Hence $$(x+y+z)(xy+yz+xz-2)\geq 9(z)(z+1)(z+2)$$here

$x=x,y=x+a,z=x+a+b$ $9x^3 + (15a + 9b)x^2 + (9a^2 + 7ab + 2b^2)x + 2a^3 + 2a^2b + ba^2 + ab^2 - 6x - 4a - 2b\ge 9xyz = 9x^3 + 9ax^2 + 9bx^2 + 9a^2x + 9abx$ $6ax^2-2abx+2b^2x+2a^3+2a^2b+ba^2+ab^2-6x-4a-2b\ge 0$ which is true

sqing wrote: Let $x,y,z$ be positive integers such that $x\neq y\neq z \neq x$ .Prove that $$(x+y+z)(xy+yz+zx-2)\geq 9xyz.$$When does the equality hold? Proposed by Dorlir Ahmeti, Albania Let $a,b,c,d\in\mathbb{Z}_{\geq0}$ be pairwise distinct. Prove $$\left(\sum{a}\right)\left(-5+\sum{ab}\right)\geq6\left(abc+abd+acd+bcd\right),$$where $\sum{ab}=ab+ac+ad+bc+bd+cd.$