Lax entropy condition for convex flux and sign of shock speed 
Assume that $u$ is an entropy solution to $u_t+\partial_x(A(u))=0$, where $A$ is a non-decreasing function on $\mathbb{R}$ such that $A(0)=0$ and $$A'(u_\ell)>\gamma'(t)=\frac{A(u_\ell)-A(u_r)}{u_\ell-u_r}>A'(u_r)$$ (namely the Lax' entropy condition, where $u_\ell, u_r$ are of the usual meanings). Prove that the shocks travel from left to right if $A$ is convex.

I got frustrated after spending a long time on it (not sure if one is supposed to consider the shock speed $\gamma'(t)$, which is always non-negative since $A$ is non-decreasing). I believe that I missed something here and that there is some very simple way to solve this problem. Can someone help me? Thanks.

The problem statement is slightly misleading. Indeed, it is assumed that the flux $A$ is convex and increasing. The speed of a shock wave is given by the Rankine-Hugoniot condition
$$
\gamma'(t) = \frac{A(u_\ell) - A(u_r)}{u_\ell - u_r}
$$
so as to ensure that the shock is a weak solution. Since $A$ is convex, the shock must also satisfy the Lax entropy condition
$$
A'(u_\ell) > \gamma'(t) > A'(u_\ell)
$$
to be an entropy solution. To prove that $\gamma'(t) > 0$, we use the fact that $A$ is non-decreasing. 
Therefore, admissible shocks propagate to the right if $A$ is convex and non-decreasing (note that $A(0)=0$ is not required). In the case of the inviscid Burgers' equation $A(u) = \frac{1}{2}u^2$, this is only true for $u \geq 0$.
