$x,y,z \geqslant 0$, prove
$$\frac{(x^2+y^2+z^2)^2}{x^3y+y^3z+z^3x} \geqslant 2 \left(\frac{xy^2+yz^2+zx^2}{x^2y+y^2z+z^2x} \right)+\frac{x^2y+y^2z+z^2x}{xy^2+yz^2+zx^2}$$
What I tried
Denote $u=xy^2+yz^2+zx^2$ and $v=x^2y+y^2z+z^2x$. There is a strong inequality
$$(x^2+y^2+z^2)^2\geqslant 3 (x^3y+y^3z+z^3x)$$
I attempted to prove $2\frac{u}{v}+\frac{v}{u} \leqslant 3$ but it is too difficult.
Let $x=\min\{x,y,z\}$, $y=x+u$ and $z=x+v$.
Hence, $$\left(\frac{(x^2+y^2+z^2)^2}{x^3y+y^3z+z^3x}-\frac{2(xy^2+yz^2+zx^2)}{x^2y+y^2z+z^2x} -\frac{x^2y+y^2z+z^2x}{xy^2+yz^2+zx^2}\right)\sum_{cyc}x^3y\sum_{cyc}x^2y\sum_{cyc}x^2z=$$ $$=9(u^2-uv+v^2)x^8+9(3u^3-4u^2v+3uv^2+3v^3)x^7+21(2u^4-3u^3v+4uv^3+2v^4)x^6+$$ $$+(39u^5-42u^4v-72u^3v^2+87u^2v^3+99uv^4+39v^5)x^5+$$ $$+(22u^6-2u^5v-79u^4v^2+8u^3v^3+106u^2v^4+65uv^5+22v^6)x^4+$$ $$+(7u^7+12u^6v-33u^5v^2-33u^4v^3+47u^3v^4+56u^2v^5+27uv^6+7v^7)x^3+$$ $$+(u^8+6u^7v-5u^6v^2-14u^5v^3-3u^4v^4+30u^3v^5+14u^2v^6+7uv^6+v^8)x^2+$$ $$+uv(u^7-2u^5v^2-u^4v^3+9u^2v^5+uv^6+v^7)x+u^3v^7\geq0$$ Done!
For example, by AM-GM $$u^7-2u^5v^2-u^4v^3+9u^2v^5+uv^6+v^7=$$ $$=\frac{1}{2}(u^7-4u^5v^2+16u^2v^5+u^7-2u^4v^3+2u^2v^5)+uv^6+v^7\geq$$ $$\geq\frac{u^2}{2}\left(3\cdot\frac{1}{3}u^5+2\cdot8v^5-4u^3v^2\right)+\frac{u^2}{2}\left(2\cdot\frac{1}{2}u^5+3\cdot\frac{2}{3}v^5-2u^2v^3\right)\geq$$ $$\geq\frac{u^5v^2}{2}\left(5\sqrt[5]{\left(\frac{1}{3}\right)^3\cdot8^2}-4\right)+\frac{u^4v^3}{2}\left(5\sqrt[5]{\left(\frac{1}{2}\right)^2\left(\frac{2}{3}\right)^3}-2\right)\geq0.$$