After Ravi substitution it's equivalent to (all sums are cyclic) $11\sum x^3z+33xyz\sum x\le 4\sum x^4+16\sum x^3y+ 24\sum x^2y^2$

From WolfusA, we want to show that: $$33\sum_{cyc} x^3y+ 99\sum_{cyc} x^2yz \le 12 \sum_{cyc} x^4 + 48 \sum_{cyc} x^3z+ 72 \sum_{cyc} x^2y^2.$$ We know from the Trivial Inequality that: $$12 \sum_{cyc} x^4 - 40 \sum_{cyc} x^3y + 40\sum_{cyc} x^3z + 44 \sum_{cyc} x^2y^2 - 56 \sum_{cyc} x^2yz = \sum_{cyc} 4(x^2 - y^2 + z^2 - 3xy + 2yz)^2 \ge 0.$$Hence, it would be enough to show that: $$7 \sum_{cyc} x^3y + 8 \sum_{cyc} x^3z + 28 \sum_{cyc} x^2y^2 -43 \sum_{cyc} x^2yz \ge 0.$$(1) By AM-GM, we know that $4x^3y + y^3z + 2z^3x \ge 7x^2yz,$ and so summing with cyclic variants gives that: $$7 \sum_{cyc} x^3y \ge 7 \sum_{cyc} x^2yz.$$ Analogously/by symmetry, we get that: $$8 \sum_{cyc} x^3z \ge 8\sum_{cyc} x^2yz.$$ Finally, it remains to note that: $$28 \sum_{cyc} x^2y^2 \ge 28 \sum_{cyc} x^2yz,$$ and so adding up the three previously obtained inequalities gives the desired inequality (1). $\square$

Vlados021 wrote: Let $a$, $b$, $c$ be the lengths of the sides of a triangle $ABC$. Prove that $$ a^2(p-a)(p-b)+b^2(p-b)(p-c)+c^2(p-c)(p-a)\leqslant\frac{4}{27}p^4, $$where $p$ is the half-perimeter of the triangle $ABC$. After use Ravi substitution, inequality become \[xy^3+yz^3+zx^3+3xyz(x+y+z) \leqslant \frac{4}{27}(x+y+z)^4.\]We have \[xy^3+yz^3+zx^3+3xyz(x+y+z) \leqslant xy^3+yz^3+zx^3+(xy+yz+zx)^2 = (x+y+z)(xy^2+yz^2+zx^2+xyz).\]Therefore, we need to prove \[xy^2+yz^2+zx^2+xyz \leqslant \frac{4}{27}(x+y+z)^3.\]Clearly.

Nguyenhuyen_AG wrote: Therefore, we need to prove \[xy^2+yz^2+zx^2+xyz \leqslant \frac{4}{27}(x+y+z)^3.\]Clearly. Sorry, how is this clear?

Pathological wrote: Sorry, how is this clear? It's very old problem https://artofproblemsolving.com/community/c6h100960p588429 https://artofproblemsolving.com/community/c6h1278873p6832339

Assume WLOG $x=\min\lbrace x,y,z\rbrace$. We can set $y=x+a\wedge z=x+b$ where $a,b\ge 0$. Then $4(x+y+z)^3-27(xy^2+yz^2+zx^2+xyz)=9(a^2+ab+b^2)x+5a^3-6a^2b-3ab^2+4b^3$ It's a linear non decreasing function in $x$ as $a^2+ab+b^2\ge 0$. Hence we are left to prove $0\le5a^3-6a^2b-3ab^2+4b^3=(a-b)^2(5a+4b)$ which is true

Just note that $a^2(p-a) \le \frac{4p^3}{27}$ by $AM-GM$.

Beautiful and simple! Thanks