= Solution
Let $U_s=U(y_s)$ at an <inflection point> supplied by the preceding identity. Its imaginary part, divided by nonzero $c_i$, gives $\int U''|\psi|^2/|U-c|^2dy=0$. Its real part is
$$
I+\int_{y_1}^{y_2}\frac{U''(U-c_r)}{|U-c|^2}|\psi|^2dy=0.
$$
Adding $(c_r-U_s)$ times the zero <integral> replaces $U-c_r$ by $U-U_s$:
$$
\int_{y_1}^{y_2}\frac{U''(U-U_s)}{|U-c|^2}|\psi|^2dy=-I<0.
$$
Therefore the integrand's profile factor must be strictly negative somewhere in the open channel:
$$
\boxed{U''(y)[U(y)-U(y_s)]<0\quad\text{for some }y_1<y<y_2.}
$$
This is the <Fjørtoft theorem>. The positive weight and the strictly negative <integral> establish the strict inequality, not merely a nonpositive value. It is a stronger necessary condition than <Rayleigh's inflection-point theorem>, and neither condition alone proves instability.
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