Which is bigger?

Using AM-GM we have $\frac{a^3}{bc}+b+c\geq 3a$ $\frac{b^3}{ac}+a+c\geq 3b$ $\frac{c^3}{ab}+a+b\geq 3c$ Combine all the inequalities and we get $y+2x\geq 3x$ $\Leftrightarrow y\geq x$ So the answer is " $y\geq x$ always"

Let's prove this using Muirhead's Inequality...

Let $y \ge x$ $\begin{aligned} \Rightarrow abc\cdot y &\ge abc\cdot x \\ \Rightarrow abc \bigg(\dfrac{a^3}{bc}+\dfrac{b^3}{ca}+\dfrac{c^3}{ca} \bigg) &\ge abc(a+b+c) \\ \Rightarrow a^4 + b^4 + c^4 &\ge a^2bc+ab^2c+abc^2 \\ \Rightarrow \sum_{sym} a^4 &\ge \sum_{sym} a^2bc \\ \Rightarrow M(4,0,0) &\ge M(2,1,1) \end{aligned}$ which is true since 4,0,0 majorizes 2,1,1. Thus our first guess that $y \ge x$ is correct.

WLOG, let $a\leq b\leq c$
Then by rearrangement inequality we have
$y=\dfrac{a^3}{bc}+\dfrac{b^3}{ca}+\dfrac{c^3}{ab}$
$\geq\dfrac{a^3}{ca}+\dfrac{b^3}{ab}+\dfrac{c^3}{bc}$
$=\dfrac{a^2}c+\dfrac{b^2}a+\dfrac{c^2}b$
$\geq\dfrac{a^2}a+\dfrac{b^2}b+\dfrac{c^2}c$
$=a+b+c$
$=x$
From the equality case of rearrangement inequality, $y=x$ when $a=b=c$

I saw there are nice solutions to this down the comments, but I think most of us came to this as on the page of Muirhead's Inequality, so I'll give a solution using only that.

Note: we can't use negative numbers as exponents in Muirhead, so let us multiply by $abc$ . Thus, $abcx=a^2bc+b^2ac+c^2ab$ , $abcy=a^4+b^4+c^4$ We have to multiply both sides by 2 for the Muirhead to be complete (see the AM-GM part in Wiki, where it explains that for example M(2,1,1) means there will be a $a^2bc$ and a $a^2cb=a^2bc$ expression too)

$2abcx=a^2bc+a^2cb+b^2ac+b^2ca+c^2ab+c^2ba=M(2,1,1)$ $2abcy=a^4+a^4+b^4+b^4+c^4+c^4=a^4b^0c^0+a^4c^0b^0+b^4a^0c^0+b^4c^0a^0+c^4a^0b^0+c^4b^0a^0=M(4,0,0)$ . Since $M(4,0,0) \ge M(2,1,1)$ , $y \ge x$ , and the exercise is done.

$\because (3,-1,-1) \succ (1,0,0) \\\therefore\sum_{sym}\frac{a^3}{bc}\ge\sum_{sym}a\\\Leftrightarrow 2\sum_{cyc}\dfrac{a^3}{bc}\ge2(a+b+c)\\\Leftrightarrow y\ge x$

Given $y=\frac{a^3}{bc}+\frac{b^3}{ca}+\frac{c^3}{ab},$ if we give all of the fractions the same denominator, namely $abc$ , we get $=\frac{a^4}{abc}+\frac{b^4}{abc}+\frac{c^4}{abc} = \frac{a^4+b^4+c^4}{abc}.$ Using the inequality $a^2+b^2+c^2≥ab+bc+ca$ , we have $y=\frac{a^4+b^4+c^4}{abc}≥\frac{a^2b^2+b^2c^2+c^2a^2}{abc}≥\frac{ab^2c+abc^2+a^2bc}{abc}=\frac{abc(a+b+c)}{abc}=a+b+c=x.$ Thus, $y≥x.$