A question involving directional derivatives and differential inequalities This is a follow-up question to A question about copulas and directional derivatives. Since no answer was given, I am going to precise the definition of copula. I am interested in proving (or disproving) that
\begin{align*}
\langle\nabla C(a,a),(a,a)\rangle \geq C(a,a)
\end{align*}
where $0\leq a \leq 1$ and $C$ is a bivariate copula. A bivariate copula is a function $C:[0,1]^{2}\rightarrow[0,1]$ such that
\begin{align*}
\begin{cases}
C(1,t) = C(t,1) = t\\\\
C(0,t) = C(t,0) = 0\\\\
\displaystyle\frac{\partial^{2}C}{\partial u\partial v} = \frac{\partial^{2} C}{\partial v\partial u} \geq0
\end{cases}
\end{align*}
It also satisfies the properties
\begin{align*}
\begin{cases}
\max\{u+v-1,0\} \leq C(u,v) \leq \min\{u,v\}\\\\
\displaystyle 0 \leq \frac{\partial C}{\partial u} \leq 1\\\\
\displaystyle 0 \leq \frac{\partial C}{\partial v} \leq 1
\end{cases}
\end{align*}
This is all that I know for the moment. I tested the given property for $C(u,v) = \min\{u,v\}$, $C(u,v) = uv$, $C(u,v) = \frac{uv}{1-(1-u)(1-v)}$ and $C(u,v) = u + v - 1 + \theta(1-u)(1-v)$ and it has worked quite well so far. Based on such considerations, could someone provide a partial or full answer to my question? It is worthy emphasizing that it is not an exercise. It makes part of my research on the theory of copulas.
EDIT
Sorry gentlemen, but I forgot to mention that $C$ must be symmetric. 
 A: No, the inequality
$$
\langle\nabla C(a,a),(a,a)\rangle\ge C(a,a) \quad\forall a\in[0,1] 
$$
does not hold in general. 
Indeed, a copula is just the (joint) cdf of a bivariate distribution on the unit square $[0,1]^2$ with the uniform marginals. That is, a map $C\colon[0,1]^2\to\mathbb R$ is a copula iff $C(x,y)=\mu([0,x]\times[0,y])$ for all $(x,y)\in[0,1]^2$, where $\mu$ is any probability measure on $[0,1]^2$ such that $\mu([0,x]\times[0,1])=\mu([0,1]\times[0,x])=x$ for all $x\in[0,1]$. 
Let now $\mu$ be the uniform distribution on the intersection of the union of the straight lines $x+y=1/2$ and $x+y=3/2$ with the square $[0,1]^2$. This intersection is shown here: 

Let then $C$ be the corresponding copula.
$\Big($One might note at this point that for this copula $C$ we have 
$$C(x,y)=\max \left(0,\min \left(\tfrac{1}{2},x\right)+\min \left(\tfrac{1}{2},y\right)-\tfrac{1}{2}\right)+\max
   \left(0,\max \left(\tfrac{1}{2},x\right)+\max \left(\tfrac{1}{2},y\right)-\tfrac{3}{2}\right) \tag{1}
$$
for $x$ and $y$ in $[0,1]$; this is straightforward but a bit tedious -- and, in fact, unnecessary -- to establish.$\Big)$ 
Directly from the above picture or, alternatively, from (1), we can see that for any $a\in(1/2,3/4)$ we have $C(a,a)=1/2$ and hence 
$$
\langle\nabla C(a,a),(a,a)\rangle=0\not\ge1/2= C(a,a). 
$$
A: Consider $C(u,v)$ be such that


*

*On the subset $\{ u \leq 1/2\} \cup \{v \leq 1/2\}$, we set $C(u,v) = \min(u,v)$.

*On the subset $\{ u,v > \frac12\}$, set
$ C(u,v) = \max(u+v - 1, \frac12)$
(One can actually summarize this to be $C(u,v) = \max(u+v - 1, \min(\frac12, u, v))$. )
One easily checks that


*

*$C(0,v) = C(u,0) = 0$

*$C(1,v) = v$, $C(u,1) = u$. 

*$C$ is continuous across the lines $u = 1/2$ and $v = 1/2$. 

*Fixed $u,v$, the functions $s\mapsto C(u,s)$ and $s\mapsto C(s,v)$ are both monotone (but not strict). 

*Obviously the comparison $\max(u+v-1,0) \leq C(u,v) \leq \min(u,v)$ holds. 


The derivatives of $C$ satisfy 


*

*On the subset $\{u < \min(v,1/2)\}$, we have $C(u) = u$ so $C_u = 1$ and $C_v = 0$;

*similarly on $\{v < \min(u,1/2)\}$, we have $C_v = 1$ and $C_u = 0$. 

*On the subset $\{u,v > 1/2, u + v < 3/2\}$, we have that $C(u,v) = 1/2$ so that $C_u = C_v = 0$. 

*Finally, on the subset that $\{u,v > 1/2, u+v > 3/2\}$ we have that $C(u,v) = u+v - 1$ so that $C_u = C_v = 1$. 


Diagramming simply one see that $C_u$ is increasing in $v$ and $C_v$ is increasing in $u$. So except for the non-differentiability along the segments 
$$ \{(s,s) : s \in [0,1/2]\}, \{ (s,1/2) : s\in [1/2,1]\}, \{ (1/2,s) : s\in [1/2,1]\}, \{ (s,3/2-s): s \in [1/2,1] \} $$
$C$ verifies all the conditions of your copula. Along those points of non-differentiability the differential inequalities you quote are also satisfied in a weak sense so that you can smooth the function slightly to get differentiability still satisfied. 
But obviously we have that at $a = 5/8$, the directional derivative
$$ \langle \nabla C(a,a), (a,a) \rangle = 0 \not\geq \frac12 = C(a,a)$$
