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How do you order $(ab^2,bc^2,cd^2,da^2)$ ???

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Ok thanks...

lsi wrote: Don't need to. $[a \ b \ c \ d][ab^2 \ bc^2 \ cd^2 \ da^2] \leq [\sigma_1 \ \sigma_2 \ \sigma_3 \ \sigma_4][ab^2 \ bc^2 \ cd^2 \ da^2]$ (imagine the first vector is vertical) for any $\sigma$ a permutation of $\{a,b,c,d \}$. $a\le b\le c\le d$ doesn't give $ab^2\ge bc^2\ge cd^2\ge da^2$.

totode wrote: let $a\le b\le c \le d$ show that: $$ab^3+bc^3+cd^3+da^3\ge a^2b^2+b^2c^2+c^2d^2+d^2a^2$$ they can't be negative

totode wrote: let $a\le b\le c \le d$ show that: $$ab^3+bc^3+cd^3+da^3\ge a^2b^2+b^2c^2+c^2d^2+d^2a^2$$ For non-negative variables we obtain: $$\sum_{cyc}(ab^3-a^2b^2)=\sum_{cyc}ab^2(b-a)=\sum_{cyc}\left(ab^2(b-a)+\frac{1}{4}(a^4-b^4)\right)=$$$$=\frac{1}{4}\sum_{cyc}(a-b)^2(a^2+2ab-b^2)\geq$$$$\geq\frac{1}{4}\left((a-b)^2(a^2-b^2)+(b-c)^2(b^2-c^2)+(c-d)^2(c^2-d^2)+(d-c+c-b+b-a)^2(d^2-a^2)\right)\geq$$ $$\geq\frac{1}{4}\left((a-b)^2(a^2-b^2)+(b-c)^2(b^2-c^2)+(c-d)^2(c^2-d^2)+((d-c)^2+(c-b)^2+(b-a)^2)(d^2-a^2)\right)=$$$$=\frac{1}{4}\left((a-b)^2(d^2-b^2)+(b-c)^2(b^2-c^2+d^2-a^2)+(c-d)^2(c^2-a^2)\right)\geq0.$$

The following inequality is also true. Let $x_1\geq x_2\geq...\geq x_n\geq0,$ $n\geq2$, $m\geq k\geq1$. Prove that $$x_1^mx_2^k+x_2^mx_3^k+...+x_n^mx_1^k\geq x_1^kx_2^m+x_2^kx_3^m+...+x_n^kx_1^m.$$

totode wrote: let $a\le b\le c \le d$ show that: $$ab^3+bc^3+cd^3+da^3\ge a^2b^2+b^2c^2+c^2d^2+d^2a^2$$ It's from JBMO's shortlist 2017,my problem . How did it get here /

arqady wrote: The following inequality is also true. Let $x_1\geq x_2\geq...\geq x_n\geq0,$ $n\geq2$, $m\geq k\geq1$. Prove that $$x_1^mx_2^k+x_2^mx_3^k+...+x_n^mx_1^k\geq x_1^kx_2^m+x_2^kx_3^m+...+x_n^kx_1^m.$$ Not Trivial rearrangement !

didnt know it came from a shortlist, could anyone delete this!

Idea_lover wrote: Trivial rearrangement ! Show us your solution.

Didn't read the question well

@above, well, it fails for the last term.

Thanks WIzard for pointing it out

arqady wrote: The following inequality is also true. Let $x_1\geq x_2\geq...\geq x_n\geq0,$ $n\geq2$, $m\geq k\geq1$. Prove that $$x_1^mx_2^k+x_2^mx_3^k+...+x_n^mx_1^k\geq x_1^kx_2^m+x_2^kx_3^m+...+x_n^kx_1^m.$$ I invented back in 2016 one tougher and a little more general . But the editors declined it . I think too many informations . Congratulations for the result .

mihaig wrote: arqady wrote: The following inequality is also true. Let $x_1\geq x_2\geq...\geq x_n\geq0,$ $n\geq2$, $m\geq k\geq1$. Prove that $$x_1^mx_2^k+x_2^mx_3^k+...+x_n^mx_1^k\geq x_1^kx_2^m+x_2^kx_3^m+...+x_n^kx_1^m.$$ I invented back in 2016 one tougher and a little more general . But the editors declined it . I think too many informations . Congratulations for the result . Can u please post it ?

Not yet .Because the idea of solution from there I want to use in another context. As problem,is not big difference between Arqady's problem and the mentioned .

Can someone help me? Can we solve this problem with Muirhead inequality?

mihaig wrote: strong_boy wrote: Pleas check my solution . Is it true? Let $f(a,b,c,d) : ab^3+bc^3+cd^3+da^3 - (a^2b^2+b^2c^2+c^2d^2+d^2a^2) \geq 0$ we know $f(a,b,c,d) \geq f(0,0,a+b,c+d)$ . It mean : Why? I mixed my variable And I tried to set one of my variables to the bounding point of the domain

Ok. When you have time, tell more about your mixing

strong_boy wrote: mihaig wrote: strong_boy wrote: Pleas check my solution . Is it true? Let $f(a,b,c,d) : ab^3+bc^3+cd^3+da^3 - (a^2b^2+b^2c^2+c^2d^2+d^2a^2) \geq 0$ we know $f(a,b,c,d) \geq f(0,0,a+b,c+d)$ . It mean : Why? I mixed my variable And I tried to set one of my variables to the bounding point of the domain So, how do you mix variables for non symmetric functions?

Here my function is summetrixc.usually i dont use mixing for non symmetric functions My answer is falsi?

I am trying to understand this "we know $f(a,b,c,d) \geq f(0,0,a+b,c+d)$"

My answer is false

No problem. I am sure you will soon prove it

mihaig wrote: No problem. I am sure you will soon prove it

Indeed, I have trust in you. You are good

mihaig wrote: Indeed, I have trust in you. You are good Thanks !!!!! Where are you from?

From Romania

mihaig wrote: From Romania Nice to meet you !

Nice to meet you too

arqady wrote: No! Muirhead does not help here because the starting inequality is not symmetric. Of course...

The problem condition also says they are non-negative, please correct it! Time-consuming but natural - be a buffalo! With $b = a+x, c = a+x+y, d = a+x+y+z$ the difference of the two sides is $a^2 x^2 + a^2 x y + a^2 x z + a^2 y^2 + a^2 y z + a^2 z^2 + a x^3 + 3 a x^2 y + 3 a x^2 z + 4 a x y^2 + 6 a x y z + 4 a x z^2 + a y^3 + 3 a y^2 z + 4 a y z^2 + a z^3 + x^3 y + x^3 z + 2 x^2 y^2 + 3 x^2 y z + 2 x^2 z^2 + x y^3 + 3 x y^2 z + 4 x y z^2 + x z^3 + y^3 z + 2 y^2 z^2 + y z^3$ which is non-negative.