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?
Best Answer
Finding an answer by weighted AM-GM and Muirhead is very directly linked, though it may not simplify the work needed in practice, you'll have to solve a set of simultaneous linear equations in either case. Majorisation $x \succ y$ (ie. Muirhead) is equivalent to $y$ being a convex combination of $x$ and its permutations. Using your example, this means: $$(4,0,0) \succ (2, 1, 1) \iff w_1(4,0,0)+w_2(0, 4, 0)+w_3(0, 0, 4)=(2, 1, 1)$$ for some non-negative weights $w_i$ s.t. $\sum_i w_i=1$. In fact finding these $w_i$ will give you AM-GM coefficients to use immediately. In this case, it is easily seen the weights $w = (\frac12, \frac14, \frac14)$ does the job, so on cyclically summing $$\frac12a^4+\frac14b^4+\frac14c^4 \geqslant a^2bc$$ you get an equivalent AM-GM proof.