An inequality for copulas 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?
 A: Let $$C_t(u,v)=\frac{uv}{(u^t+v^t-u^t v^t)^{1/t}},$$ (that's the Clayton $t$-copula) and $$h_t(u,v)=C_t(a,a)-a^2-C_t(u,v)+uv,$$ typified by GRAPH OF h_1(u,v).
Proof that the proposed inequality holds for $t=1:$
$$h_1(u,v)=\frac{(u-v)^2[4(1-u)(1-v)+(u+v)(u+v-uv)]}{4(4-u-v)(u+v-uv)}\geq 0.$$
Can someone prove that the proposed inequality holds for $C_t(u,v)$ when $t>1$?
A: Let $t \ge 1$ and set $\theta = 1/t$. We have
\begin{equation*}
 C(u,v) = \frac{uv}{(u^t+v^t- u^tv^t)^\theta}.
\end{equation*}
We wish to prove,
\begin{eqnarray*}
  C(a,a)-a^2 &\ge& C(u,v)-uv\\
  a^2\left[\frac{1}{(2a^t-a^{2t})^\theta}-1\right] &\ge& uv\left[\frac{1}{(u^t+v^t-u^tv^t)^\theta}-1\right].
\end{eqnarray*}
Since $a^2 \ge uv$ by definition, we consider the terms in the brackets. First, notice that since $u,v \in [0,1]$, we have $u^t+v^t-u^tv^t \le 1$ and also $2a^t-a^{2t} \le 1$. Thus, the terms in the brackets are nonnegative, hence it suffices to establish
\begin{equation*}
    \frac{1}{(2a^t - a^{2t})^\theta} \ge \frac{1}{(u^t+v^t - u^tv^t)^\theta}. 
\end{equation*}
But this is true since due to convexity of $x^t$ for $t \ge 1$ and AM-GM inequality we have
\begin{eqnarray*}
 a^t &=& \left(\tfrac{u+v}{2}\right)^t \le \tfrac{u^t+v^t}{2},\quad -a^{2t} \le -u^tv^t\\
 \implies &&2a^t - a^{2t} \le u^t+v^t - u^tv^t,
\end{eqnarray*}
so that taking reciprocals and using monotonicity of $x^\theta$ we are done.
