How does one even approach this problem? Convexity, uvw, Holder's, AM-GM, Homgenization all seem useless in this situation. Is it $a+b+c+d=1$ maybe?

wanwan4343 wrote: Let $a,b,c,d$ be any real numbers such that $a+b+c+d=0$, prove that \[1296(a^7+b^7+c^7+d^7)\le637(a^2+b^2+c^2+d^2)^7\] Wron g. Choose as counter-example $(a,b,c,d)=\left(-\frac 1{10},-\frac 1{10},-\frac 1{10},\frac 3{10}\right)$

the question is 1296(a^7+b^7+c^7+d^7)^2

shiangge wrote: the question is 1296(a^7+b^7+c^7+d^7)^2 I had edited 3 days ago!

wanwan4343 wrote: Let $a,b,c,d$ be any real numbers such that $a+b+c+d=0$, prove that \[1296(a^7+b^7+c^7+d^7)^2\le637(a^2+b^2+c^2+d^2)^7\] $uvw$ kills it! $62208(a^7+b^7+c^7+d^7)^2\le8281(a^2+b^2+c^2+d^2)^7$ is also true (with the same condition).

shiangge wrote: the question is 1296(a^7+b^7+c^7+d^7)^2 The question has been changed after my answer to the first version. Check dates.

arqady wrote: wanwan4343 wrote: Let $a,b,c,d$ be any real numbers such that $a+b+c+d=0$, prove that \[1296(a^7+b^7+c^7+d^7)^2\le637(a^2+b^2+c^2+d^2)^7\] $uvw$ kills it! $62208(a^7+b^7+c^7+d^7)^2\le8281(a^2+b^2+c^2+d^2)^7$ is also true (with the same condition). How to use UVW to this problem?

$d=-a-b-c$, $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.

Please expand on the above uvw solution.

arqady wrote: $d=-a-b-c$, $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$. so....how to solve it ? is there other way to solve this problem ? ( i don't know UVW very much )

It reduces to $81(S_2^2S_3-S_3S_4)^2 \le 13 P_2^7$ which I have no clue how to solve.

What about $(a+b+c+d)(a^{6} +b^{6}+ c^{6} +d^{6}) = (a^7 + b^7 + c^7 + d^7)+\frac{1}{2} \sum_{sym}a^{6}b$