Suppose that $f$ from $[0,\infty]$ onto $[0,1]$ is completely monotonic on $(0,\infty)$, and let $g$ be the inverse of $f$. For $(u,v)$ in $[0,1]^{2}$, define $C(u,v) = f(g(u)+g(v))$, and let $a = (u+v)/$2. Is $$C(a,a) - a^{2} \geq C(u,v) - uv$$ for all $(u,v)$ in $[0,1]^{2}$?

Some background: $C$ is an Archimedean copula, and if $f(x) = \mathrm{e}^{-x}$, then $C(u,v) = uv$. The complete monotonicity of $f$ is necessary and sufficient for $C$ to be extendible to a multivariate Archimedean copula, for the purpose of "joining" arbitrarily many distribution functions to form a joint distribution. Well-known (families of) copulas which apparently satisfy the inequality are the Clayton, Frank, and Gumbel families. A possibly new family (for $t>0$) results from $$f(x) = (\mathrm{e}^{(x+1)^{-t}}-1)/(e-1)$$.

"Completely monotonic" is not another term for "strictly monotonic"; $f$ is completely monotonic on (a,b) if $$(-1)^{k}f^{(k)}(x) \geq 0$$ for all x in (a,b). (Widder, *The Laplace Transform*, 1946, p. 145)

May 26 2015 - the question remains largely **unanswered.** The two "Answers" have proofs for simple examples, but the question is for a very large class of copulas (see, for example Roger Nelsen, *An Introduction of Copulas, 2nd ed.*, pp. 151-155). Can someone find a proof that uses the complete monotonicity of $f$ - or a counterexample?

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