Suppose that $X$ and $Y$ are random variables in $L^p$, $p>2$ and suppose that for Lipschitz functions $f$ and $g$,
$$
|\operatorname{Cov}(f(X),g(Y))|\le \big(\operatorname{Lip}(f)\operatorname{Lip}(g)+\|X\|_p\|Y\|_p\big)\alpha^{1-\frac{2}{p}}.\tag{1}\label{1}
$$
When $f$ and $g$ are the identity functions, the inequality becomes
$$
|\operatorname{Cov}(X,Y)|\le \big(1+\|X\|_p\|Y\|_p\big)\alpha^{1-\frac{2}{p}}.\tag{2}\label{2}
$$
Now, multiplying $X$ and $Y$ by a constant $c>0$, we get
$$
|\operatorname{Cov}(X,Y)|\le \inf_{c>0}\big(c^{-2}+\|X\|_p\|Y\|_p\big)\alpha^{1-\frac{2}{p}}=\|X\|_p\|Y\|_p\alpha^{1-\frac{2}{p}}.\tag{3}\label{3}
$$
This result seems weird because the inequality in \eqref{1} is clearly "scale invariant", that is, applying \eqref{1} directly (with $f(x)=g(x)=cx$) yields
$$
|\operatorname{Cov}(cX,cY)|\le c^2\big(1+\|X\|_p\|Y\|_p\big)\alpha^{1-\frac{2}{p}}.
$$
On the other hand, applying \eqref{1} to $Z=cX$ and $W=cY$ (with $f(x)=g(x)=x$), one gets
$$
|\operatorname{Cov}(Z,W)|\le \big(1+\|Z\|_p\|W\|_p\big)\alpha^{1-\frac{2}{p}},
$$
which implies that
$$
|\operatorname{Cov}(X,Y)|\le \big(c^{-2}+\|X\|_p\|Y\|_p\big)\alpha^{1-\frac{2}{p}},
$$
and we're back to \eqref{3}.

Is the inequality in Equation \eqref{3} actually correct?