An stronger inequality than in AoPS. For $x,y,z >0.$ Prove$:$
$$\sum {\frac {y+z}{x}}+{\frac {1728 {x}^{ 3}{y}^{3}{z}^{3}}{ \left( x+y \right) ^{2} \left( y+z \right) ^{2} \left( z+x \right) ^{2} \left( x+y+z \right) ^{3}}} \geqslant 4\sum {\frac {x}{y+z }}+1$$
I check when $xyz=0$ and $x=y$ and see it's true. So I guess it's true.
So I try to get it in $uvw$ form as follow$:$
$$-26244{u}^{7}{v}^{2}{w}^{3}+19683{u}^{6}{v}^{6}+2916{u}^{6}{w}^{6}+4374{u}^{5}{v}^{4}{w}^{3}-2673{u}^{4}{v}^{2}{w}^{6}+216{u}^{3}{w}^{9}+1728{w}^{12} \geqslant 0$$
Then I don't know how to end proof for it. BW does not help here.
I'm not sure about this inequality. I found it when I prove this inequality.
Help me please. Thanks for a real lot!

$uvw$ helps!
Let $x+y+z=3u$, $xy+xz+yz=3v^2$, where $v>0$,  and $xyz=w^3$.
Thus, we need to prove that:
$$\frac{9uv^2-3w^3}{w^3}+\frac{1728w^9}{(9uv^2-w^3)^2\cdot27u^3}\geq\frac{4(27u^3-27uv^2+3w^3+9uv^2-3w^3+3w^3)}{9uv^2-w^3}+1$$ or
$$\frac{9uv^2}{w^3}+\frac{64w^9}{u^3(9uv^2-w^3)^2}\geq\frac{4(27u^3-9uv^2+2w^3)}{9uv^2-w^3}$$ or $f(w^3)\geq0,$ where
$$f(w^3)=64w^{12}+8u^3w^9+108u^6w^6-99u^2v^4w^6-972u^7v^2w^3+162u^5v^4w^3+729u^6v^6.$$
But since by Maclaurin $$u\geq v\geq w,$$ we obtain: $$f'(w^3)=256w^9+24u^3w^6+216u^6w^3-198u^2v^4w^3-972u^7v^2+162u^5v^4<0,$$
which says that $f$ decreases and it's enough to prove our inequality for a maximal value of $w^3$,
which happens for equality case of two variables.
Since our inequality is homogeneous and symmetric, it's enough to assume $y=z=1,$ which gives:
$$(x-1)^2(x-2)^2(x^4+10x^3+37x^2+20x+4)\geq0$$ and we are done!
Now we see, why BW does not help: the equality occurs also for $(x,y,z)=(2,1,1).$
