It is a common sentiment among people involved with maths Olympiads that Muirhead's inequality should only be used to find if an AM-GM proof exists.
A common bruteforce technique with inequalities is to clear denominators, multiply everything out, and apply Muirhead's or Schur's. However, it is worth noting that any inequality that can be proved directly with Muirhead can also be proved using the Arithmetic Mean-Geometric Mean inequality. In fact, IMO gold medalist Thomas Mildorf says it is unwise to use Muirhead in an Olympiad solution; one should use an application of AM-GM instead. Thus, it is suggested that Muirhead be used only to verify that an inequality can be proved with AM-GM before demonstrating the full AM-GM proof.
My question is: Can that fact guide us in finding that AM-GM proof? Other than the fact that it implies its existance.
I was trying to solve $\displaystyle a+b+c \leq \frac{a^3}{bc}+\frac{b^3}{ac}+\frac{c^3}{ab}$. By clearing denominators $\displaystyle a^2bc+ab^2c+abc^2\leq a^4+b^4+c^4$ we can get a proof by Muirhead's since $(2,1,1)\prec (4,0,0)$. But I had a harder time finding that AM-GM proof $\displaystyle \sum a^2 bc=\sum (a^3)^{2/3}(b^3)^{1/3}(c^3)^{1/3}\leq \sum 2a^3/3+b^3/3+c^3/3$. I used this example as a reference if needed.
Could the Muirhead solution guide you to how you should apply AM-GM on the cleared denominators? Maybe by the weighted AM-GM? Or does the fact that a Muirhead proof exists just mean that you should randomly look for the AM-GM proof either on the initial inequality or its cleared denominators version or even an intermediate step without giving you any further indication on how to apply it?
From weighted AM-GM, we get $$ \frac12 a^4 + \frac14 b^4 + \frac14 c^4 \ge (a^4)^{1/2} (b^4)^{1/4} (c^4)^{1/4} = a^2bc. $$ If we add up the symmetric versions of this inequality, we get exactly the inequality $$a^4 + b^4 + c^4 \ge a^2 bc + a b^2 c + a b c^2.$$
In general, whenever one exponent sequence majorizes another, we can somehow group up the terms of Muirhead's inequality and apply AM-GM several times, instead.
The grouping is not immediate, and often it is not unique. To give another example: suppose we applied Muirhead's inequality to get $$a^3b + a^3c + b^3a + b^3c + c^3a + c^3b \ge 2a^2 bc + 2a b^2 c + 2a b c^2.$$ To deduce the same inequality with AM-GM, the term $a^2bc$ must somehow be written as $$(a^3b)^{n_1} (a^3c)^{n_2} (b^3a)^{n_3} (b^3c)^{n_4} (c^3a)^{n_5} (c^3b)^{n_6}$$ where $n_1 + n_2 + n_3 + n_4 + n_5 + n_6 = 1$. This is underconstrained. For example, we can say $$\frac14 a^3 b + \frac14 a^3 c + \frac14 b^3a + \frac14 c^3 a \ge (a^3b)^{1/4} (a^3c)^{1/4} (b^3a)^{1/4} (c^3a)^{1/4} = a^2bc$$ or we can say $$\frac13 a^3b + \frac13 a^3c + \frac16 b^3c + \frac16 c^3b \ge (a^3b)^{1/3} (a^3c)^{1/3} (b^3c)^{1/6}(c^3b)^{1/6} = a^2bc.$$ For help coming up with these: I first figured that we'd need to use the $a^3b$ and/or the $a^3c$ terms to get a factor $a^2$ on the right. Since $a^2bc$ has equal exponents of $b$ and $c$, I decided we might as well treat $b$ and $c$ in the same way on the left. This is still underconstrained, so to simplify our lives, I threw some terms away to get a unique solution - in two different ways.
Whichever way we use AM-GM, once we add up all symmetric versions of the inequality, we should get the same inequality that Muirhead had.
A proof of Muirhead's inequality can be used to "chase down" one particular construction of the exponents used in weighted AM-GM, but it's probably not worth it.