I was thinking about Muirhead-style inequalities, and read that Muirhead is typically used to motivate AM-GM proofs rather than being used on their own. I tried random sequences of powers, and came up with this: Prove $x^5y^2z+xy^5z^2+x^2yz^5 \geq x^4y^3z+xy^4z^3+x^3yz^4$ for positive $x,y,z$.
This is similar to Muirhead (except it is a cyclic sum rather than symmetric), so we would expect AM-GM to work. After playing around with the powers, I ended up with $\sum_{cyc} \frac{10x^5y^2z+4xy^5z^2-x^2yz^5}{13}$ - all well and good, as the powers work out, except I used a negative, which means AM-GM is invalid. IS it possible to prove this result with just AM-GM?
Hint: Arguably the best method for proving cyclic, homogeneous inequalities of three variables is to reduce it to two variables - assume $$x = \min\{x,y,z\}$$ and write $$y = x+t, z = x+p.$$ This will yield a homogeneous inequality in three variables $x,t,p$, where there will be lots of cancellations of powers of $x$ usually. The resulting inequality of three variables is usually workable using some calculus.
I am not sure who exactly, but one of Vasile Cirtoaje, Pham Kim Hung studies these types of inequalities in detail in their famous inequality books. You can easily google search for those texts.
EDIT 1:
Whether it is possible or not to prove it using only AM-GM, I am $99$% confident that you cannot based on my experience. The best case scenario - almost always you end up having to prove some inequality $f(r)\geq 0$, where $f$ is of degree higher than $2.$ Until this point, we have only transformed the inequality without strengthening or loosening with another inequality and so a purely AM-GM solution is highly improbable.
EDIT 2:
As the below answer provides, there is a counter-example. However, the inequality remains true if you assume $x,y,z$ are sides of a triangle, in which case you can prove your inequality after doing the Ravi substituion: $$x = a+b, y=b+c, z = c+a,\,\,\,\, a,b,c>0.$$