$x,y,z>0$ and $x+y+z=1$. $x,y,z>0$ and $x+y+z=1$. Prove:
$$\sum_{cyc}\frac{x^{5}+y^{4}z}{y+z}\geq \frac{1}{27}$$
My tried was to add $3$ and to homogenize with $x + y + z$. Another tried was to write LHS as $\sum_{cyc} \frac{x^{5}}{y+z}+\sum _{cyc}\frac{y^{4}z}{y+z}$.
 A: Hint: Split the terms as
$$ \sum \frac{ x^5 - y^5 } { y+z } + \sum \frac{ y^5 + y^4 z } { y+z} \geq \frac{1}{27}.$$

 Show that
 1. the first summation is $ \geq 0$, (If stuck, see comments)
 2. the second summation is $ \sum y^4 \geq \frac{1}{27}$.


I'm not quite satisfied with this solution because the splitting seems "magical", though I know what steps / similar problems motivated the approach.
I would like to see another distinct-enough (non brute-force) approach.
The issue with $ \sum \frac{ y^4 z } { y+ z }$ is that because of the cross terms, it could get too low (even close to 0, like with $ x =y = \epsilon \rightarrow 0 , z = 1 - 2 \epsilon$).
In the event this happens, the $ \sum \frac{ x^5 } { y+z }$ term would have gotten large, and so we need to "share" it across the split.
Adding a $y^5$ term in the numerator of the second term helps us to factor it, leaving a no-cross-term $y^4$ that makes the split ultimately work.
We then have to subtract a $y^5$ term in the numerator of the first term, and check if it truly satisfies the desired bound.
