In one of my lectures, we dealt with the Hopf-Lax formula for the solution of the Hamilton-Jacobi equation with a free particle:
$$\partial_t u(t,x) + \frac{1}{2} \lvert \nabla u(x,t) \rvert^2=0 $$
We assume that the initial condition at some $t_0>0$ is globally Lipschitz and take some point $(x,t_0+s)$, $x \in \mathbb{R^d}, s>0$ where the time and space derivatives exists. For this point, we define $ \eta = (x-y)/s$ where $y$ is a minimising point in the Hopf-Lax formula, which means
$$\min_{z \in \mathbb{R^d}} \bigg\{ \frac{|x-z|^2}{2s} + u^0(z) \bigg\}= \frac{|x-y|^2}{2s} + u^0(y)$$
We then show that
$$\partial_{t} u(t,x) + \eta \cdot \nabla u(t,x) - \frac{1}{2}|\eta|^2\geq 0$$
Now the claim is that the LHS is maximised for $\eta = \nabla u(t,x).$ Why? The term $\eta \cdot \nabla u(t,x)$ is maximised for $\eta = \nabla u(t,x)$ due to Cauchy-Schwarz, but why does it hold with this additional negative term $\frac{1}{2}|\eta|^2?$
For $\mathbf{a}\cdot\mathbf{b}-a^2/2$ to be maximized, $\mathbf{a}$ must be parallel to $\mathbf{b}$. So we have that $\mathbf{a}\cdot\mathbf{b} = ab$. Thus we are trying to maximize $a(b-a/2)$ with respect to $a$, which occurs at $a=b$. Since $\mathbf{a}$ and $\mathbf{b}$ are parallel with the same magnitude, they are equal.