Two Inequality Challenges involving $\sum_{cyc}\left(\frac{P(a,b,c)}{Q(a,b,c)}\right)^r$ for homogeneous polynomials $P$ and $Q$ of equal degree Let $F(a,b,c,r)=(P(a,b,c)/Q(a,b,c))^r$ where $P(a,b,c)$ and $Q(a,b,c)$  are homogenous polynomials with equal degrees $deg(p)=deg(q)=n$, defined for non-negative $a,b$, and $c$. 
Below, we assume everywhere (for simplicity) that $a+b+c=3$.
In his "Secrets of inequalities" (vol.2, p.323) Pham Kim Hung proves a result equivalent to the folloiwng: 
If $P$ and $Q$ are linear ($n=1$) and $r=1$ then the minimum of $F(a,b,c)+F(b,c,a)+F(c,a,b)$ is achieved either at (1,1,1) or at the boundary of the region (e.g. $c=0$).
I would like to determine under what conditions this can be generalized for other values of $n$ and $r$. If you look at the Inequality tag here on StackExchange, you'll find lots of examples of inequalities which reduce to this  cyclic sum for some well-chosen $F$ and for which the extremum is at $(1,1,1)$ or the boundary. This is of course very much inderstandable given Lagrange multiplier theorem.
Here is an example ($a+b+c=3$):$$
\left(\frac{a^2 + b c}{b^2 + a c}\right)^r + \left(\frac{b^2 + a c}{a b + c^2}\right)^r + \left(\frac{a b + c^2}{a^2 + b c}\right)^r \geq 3$$
The minimum in this case is achieved both at $(1,1,1)$ and at $(1.5,1.5,0)$
I would like to propose two challenges.
Challenge 1
Find an example of $P$ and $Q$ with lowest possible degree, for which the minimum of $F(a,b,c)+F(b,c,a)+F(c,a,b)$ for $r=1$ is achieved in a non-trivial point - meaning not $(1,1,1)$ and not $abc=0$.
Challenge 2
Find an example of $P$ and $Q$ with lowest possible degree, for which the minimum of $F(a,b,c,1)+F(b,c,a,1)+F(c,a,b,1)$ is achieved in a trivial point, but there exists $r>0$ for which the minimum of $F(a,b,c,r)+F(b,c,a,r)+F(c,a,b,r)$ is achieved in a non-trivial point.

Update 08/18/2019
For challenge 1: Let
$$F(a, b, c) = \frac{a^2b + 2a^2c + 2ab^2 + b^3 + 31abc}{(a+b+50c)(a+b+c)^2}.$$
Let $G(a,b,c) = F(a, b, c) + F(b, c, a) + F(c, a, b)$.
Then the minimum of $G(a, b, c)$ is not achieved at $(1, 1, 1)$ or $abc=0$.
Indeed, we have $G(1,1,1) = 37/156 \approx 0.2372$ and
$$G(a, 3-a, 0) = \frac{49a^4-8094a^3+45900a^2-66177a-4050}{9(49a+3)(49a-150)} > 0.21, \quad \forall 0\le a\le 3.$$
However, $G(1/2, 1/8, 19/8) = 1018835/4907936 \approx 0.2076$;
For challenge 2: Let
$$F(a, b, c) = \frac{a^2b + 2a^2c + 2ab^2 + b^3 + 10 abc}{(a+b+100c)(a+b+c)^2}.$$
Let $r = \frac{1}{3}$.
The minimum of $F(a,b,c) + F(b, c, a) + F(c, a, b)$ is achieved at $(1,1,1)$. This can be proved by the Buffalo Way. 
The proof is omitted here. I also use Mathematica Resolve to verify its truth.
On the other hand, the minimum of $H(a,b,c) = F(a,b,c)^r + F(b, c, a)^r + F(c, a, b)^r$ is not achieved at $(1, 1, 1)$ or $abc = 0$.
Indeed, we have $H(1,1,1) \approx 0.7778$ and 
$H(a, 3-a, 0) > 0.9$ for $0\le a\le 3$; However, $H(5/6, 9/20, 103/60) \approx 0.7749$;
