I just began learning the uvw method and tried solving this problem using it.
I would like to get my work reviewed and my mistakes pointed out.
Prove,whenever $x,y,z \geq 0$ and $x+y+z=3$ :$$\frac{x}{y^2}+\frac{y}{z^2}+\frac{z}{x^2}\ge x^3+y^3+z^3 $$
Proof:
Homogenising gives us: $$(x+y+z)^4 \sum_{cyc}x^3y^2 \geq 81(xyz)^2\sum_{cyc}x^3$$
Now let $a+b+c=3u$ ; $ab+ac+bc=3v^2$ ; $abc=w^3$
Then the inequality is equivalent to: $$\frac{81}{2}u^4(27uv^4-18u^2w^3-3v^2(x-y)(x-z)(y-z) \geq 81w^6(27u^3-27uv^2+3w^3)$$ $\Longleftrightarrow$ $$27u^5v^4-54w^6(u^3-uv^2)-6w^9-18u^2w^3 \geq 3v^2u^4(x-y)(x-z)(y-z)$$ $\Longleftrightarrow$ $$(27u^5v^4-54w^6(u^3-uv^2)-6w^9-18u^2w^3)^2 \geq 3^5v^4u^8(-(w^3-3uv^2+2u^3)^2+4(u^2-v^2)^3)$$
Fixing $u$ , $v^2$ , this is equivalent to
$f(w^3) \geq 0$ with $$f(w^3)=36w^{18}+729u^{10}v^8+648w^{15}u^3-648uw^{15}v^2+108w^{12}u^2+108u^2w^{12}+324u^4w^6-162u^5v^4w^9-486u^7v^4w^3+2916w^{12}u^6-5832w^{12}u^4v^2+2916w^{12}u^2v^4+972w^9u^5-972w^9u^3v^2-162w^9u^5v^4-486u^7w^3v^4+972u^5w^9-972u^3w^9v^2-1458u^8v^4w^6+1458u^6v^6w^6-1458w^6u^8v^4+1458w^6u^6v^6+243v^4u^8\left(w^3-3uv^2+2u^3\right)^2-972v^4u^{14}+2916v^6u^{12}-2916v^8u^{10}+972v^{10}u^8$$
Since f is concave then the minimum happens at the extreme points and so $w^3$ assumes minimum either when two of $x,y,z$ are equal or when one of them are zero.
Case:1
$z=0$
Indeed, We have to prove $$(x+y)^4(y^3x^2)\geq 0$$
Which is obvious:
Case:2
$z=y=1$
Indeed , we have to prove :$$(x+2)^4(x^3+x^2+1)\geq 81x^2(x^3+2)$$ $\Longleftrightarrow$ $$(x-1)^2(x^5+11x^4-28x^3-10x^2+64x+16)\geq 0$$ $\Longleftrightarrow$ $$(x^5+11x^4-28x^3-10x^2+64x+16)\geq 0$$
Which is true for any $x > 0$