[proofplan]
We prove both directions of the equivalence separately. For the reverse direction ($\Leftarrow$), we assume the variational inequality $(u - v, w - u)_V \ge 0$ for all $w \in K$ and expand $\|v - w\|_V^2$ to show $\|v - w\|_V \ge \|v - u\|_V$ for every $w \in K$. For the forward direction ($\Rightarrow$), we assume $u = P_K(v)$ and use the fact that convex combinations $(1-t)u + tw$ remain in $K$ for $t \in (0,1]$; minimality of $u$ forces the derivative of $t \mapsto \|v - u - t(w-u)\|_V^2$ at $t = 0$ to be nonneg, which yields the variational inequality.
[/proofplan]
[step:Prove that the variational inequality implies $u = P_K(v)$]
Assume $u \in K$ satisfies $(u - v, w - u)_V \ge 0$ for all $w \in K$. For any $w \in K$, expand:
\begin{align*}
\|v - w\|_V^2 &= \|(v - u) + (u - w)\|_V^2 \\
&= \|v - u\|_V^2 + 2(v - u, u - w)_V + \|u - w\|_V^2.
\end{align*}
Since $(v - u, u - w)_V = -(u - v, u - w)_V = (u - v, w - u)_V \ge 0$ by hypothesis, and $\|u - w\|_V^2 \ge 0$:
\begin{align*}
\|v - w\|_V^2 \ge \|v - u\|_V^2.
\end{align*}
This holds for all $w \in K$, so $u$ minimises $\|v - \cdot\|_V$ over $K$. By uniqueness of the projection ([Existence and Uniqueness of the Projection Operator](/theorems/86)), $u = P_K(v)$.
[guided]
We want to show that the variational inequality forces $u$ to be the nearest point in $K$ to $v$. The idea is to compare $\|v - w\|_V^2$ with $\|v - u\|_V^2$ for an arbitrary $w \in K$ by inserting $\pm u$.
Write $v - w = (v - u) + (u - w)$ and expand:
\begin{align*}
\|v - w\|_V^2 &= \|v - u\|_V^2 + 2(v - u, u - w)_V + \|u - w\|_V^2.
\end{align*}
The cross term $(v - u, u - w)_V$ can be rewritten. Using linearity of the inner product:
\begin{align*}
(v - u, u - w)_V = -(u - v, u - w)_V = (u - v, w - u)_V.
\end{align*}
The hypothesis gives $(u - v, w - u)_V \ge 0$, so $(v - u, u - w)_V \ge 0$. Combined with $\|u - w\|_V^2 \ge 0$:
\begin{align*}
\|v - w\|_V^2 \ge \|v - u\|_V^2.
\end{align*}
Since $w \in K$ was arbitrary, $u$ achieves the infimum of $\|v - w\|_V$ over $K$. By the [uniqueness of the projection](/theorems/86), $u = P_K(v)$.
[/guided]
[/step]
[step:Prove that $u = P_K(v)$ implies the variational inequality]
Assume $u = P_K(v)$. Let $w \in K$ be arbitrary. Since $K$ is convex, for each $t \in (0,1]$ the convex combination
\begin{align*}
w_t := u + t(w - u) = (1-t)u + tw
\end{align*}
belongs to $K$. Define the function
\begin{align*}
\phi: [0,1] &\to \mathbb{R}, \\
t &\mapsto \|v - u - t(w - u)\|_V^2.
\end{align*}
Since $u$ minimises $\|v - \cdot\|_V$ over $K$ and $w_t \in K$ for all $t \in [0,1]$, we have $\phi(t) \ge \phi(0)$ for all $t \in [0,1]$.
Expanding $\phi(t)$:
\begin{align*}
\phi(t) &= \|v - u\|_V^2 - 2t(v - u, w - u)_V + t^2\|w - u\|_V^2.
\end{align*}
The condition $\phi(t) \ge \phi(0)$ gives
\begin{align*}
-2t(v - u, w - u)_V + t^2\|w - u\|_V^2 \ge 0.
\end{align*}
Dividing by $t > 0$:
\begin{align*}
-2(v - u, w - u)_V + t\|w - u\|_V^2 \ge 0.
\end{align*}
Taking $t \to 0^+$:
\begin{align*}
(v - u, w - u)_V \le 0.
\end{align*}
Equivalently, $(u - v, w - u)_V \ge 0$, which gives $(u, w - u)_V \ge (v, w - u)_V$.
[guided]
The forward direction uses a standard first-order optimality argument for constrained minimisation over a convex set.
Since $K$ is convex and $u, w \in K$, the line segment $\{u + t(w-u) : t \in [0,1]\}$ lies in $K$. The point $u = P_K(v)$ minimises $\|v - \cdot\|_V^2$ over $K$, so the function
\begin{align*}
\phi(t) = \|v - u - t(w-u)\|_V^2
\end{align*}
has a minimum at $t = 0$ over $[0,1]$. This is a quadratic polynomial in $t$:
\begin{align*}
\phi(t) = \|v - u\|_V^2 - 2t(v - u, w - u)_V + t^2\|w - u\|_V^2.
\end{align*}
The minimality at $t = 0$ means the right derivative $\phi'(0^+) \ge 0$. Computing:
\begin{align*}
\phi'(t) = -2(v-u, w-u)_V + 2t\|w-u\|_V^2,
\end{align*}
so $\phi'(0) = -2(v-u, w-u)_V \ge 0$, giving $(v-u, w-u)_V \le 0$.
Equivalently:
\begin{align*}
(u - v, w - u)_V \ge 0 \quad \iff \quad (u, w-u)_V \ge (v, w-u)_V.
\end{align*}
This is the variational inequality. Since $w \in K$ was arbitrary, the proof is complete.
[/guided]
[/step]
