= Solution
Keep the same function $V$. For arbitrary $a,b$ its <orbital derivative> is
$$
\dot V=r\{a x(x-1)-b y(y-1)\}.
$$
Let $\gamma$ be a nonconstant orbit of the unperturbed system with period $T$. Since
$$
\frac{\dot y}{ry}=x-1,
\qquad
\frac{\dot x}{x}=1-y,
$$
integration over one period gives
$$
\int_0^T(1-x)\,dt=0,
\qquad
\int_0^T(1-y)\,dt=0.
$$
Consequently,
$$
\begin{aligned}
A&:=\int_0^T x(x-1)\,dt
=\int_0^T(1-x)^2\,dt>0,\\
B&:=\int_0^T y(y-1)\,dt
=\int_0^T(1-y)^2\,dt>0.
\end{aligned}
$$
If a perturbed periodic orbit persisted and converged to $\gamma$ as $(a,b)\to(0,0)$, its net change in $V$ over one period would vanish. The <averaged first-integral obstruction to persistence of a periodic orbit> would therefore require
$$
0=r(aA-bB)+o(|a|+|b|).
$$
When $ab<0$, the two leading terms have the same nonzero sign and their magnitude is bounded below by $r\min(A,B)(|a|+|b|)$. \b[Thus no unperturbed periodic orbit persists for sufficiently small $a,b$ with $ab<0$.]
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