[proofplan] We identify the vector field $z \mapsto |z|^{p-2}z$ as the gradient of the convex potential $z \mapsto |z|^p/p$. Along the line segment from $\eta$ to $\xi$, this reduces the desired inequality to monotonicity of the derivative of a one-variable convex function. The strict equality statement follows because the second derivative along every non-constant line segment is strictly positive except possibly at one point. [/proofplan] [step:Define the potential and compute its gradient] Define the potential function $F: \mathbb{R}^n \to \mathbb{R}$ by \begin{align*} F(z) = \frac{1}{p}|z|^p. \end{align*} Define the vector field $A: \mathbb{R}^n \to \mathbb{R}^n$ by $A(0)=0$ and, for $z \neq 0$, \begin{align*} A(z) = |z|^{p-2}z. \end{align*} We claim that $F$ is differentiable on $\mathbb{R}^n$ and $\nabla F(z)=A(z)$ for every $z \in \mathbb{R}^n$. If $z \neq 0$, the [Chain Rule for Maps Between Euclidean Spaces](/theorems/323) applied to $z \mapsto |z|^2$ and $s \mapsto s^{p/2}/p$ gives \begin{align*} \nabla F(z) = |z|^{p-2}z = A(z). \end{align*} At $z=0$, for $h \in \mathbb{R}^n \setminus \{0\}$, \begin{align*} \frac{|F(h)-F(0)-A(0)\cdot h|}{|h|} = \frac{|h|^{p-1}}{p}. \end{align*} Since $p>1$, the right-hand side tends to $0$ as $h \to 0$. Hence $F$ is differentiable at $0$ and $\nabla F(0)=0=A(0)$. [/step] [step:Restrict the potential to the line segment from $\eta$ to $\xi$] Fix $\xi,\eta \in \mathbb{R}^n$. Define $v \in \mathbb{R}^n$ by \begin{align*} v = \xi-\eta. \end{align*} Define the line segment map $\gamma: [0,1] \to \mathbb{R}^n$ by \begin{align*} \gamma(t) = \eta + tv. \end{align*} Define the one-variable function $\varphi: [0,1] \to \mathbb{R}$ by \begin{align*} \varphi(t) = F(\gamma(t)) = \frac{1}{p}|\eta + tv|^p. \end{align*} Since $F$ is differentiable and $\gamma$ is affine, the [Chain Rule for Maps Between Euclidean Spaces](/theorems/323) gives, for every $t \in [0,1]$, \begin{align*} \varphi'(t) = A(\gamma(t))\cdot v. \end{align*} In particular, \begin{align*} \varphi'(1)-\varphi'(0) = \big(A(\xi)-A(\eta)\big)\cdot(\xi-\eta). \end{align*} [guided] The purpose of this step is to reduce the vector inequality to a one-dimensional monotonicity statement. We fix $\xi,\eta \in \mathbb{R}^n$ and define the displacement vector $v \in \mathbb{R}^n$ by \begin{align*} v = \xi-\eta. \end{align*} The affine path from $\eta$ to $\xi$ is the map $\gamma: [0,1] \to \mathbb{R}^n$ given by \begin{align*} \gamma(t) = \eta + tv. \end{align*} Thus $\gamma(0)=\eta$ and $\gamma(1)=\xi$. Now define $\varphi: [0,1] \to \mathbb{R}$ by restricting the potential $F$ to this segment: \begin{align*} \varphi(t) = F(\gamma(t)) = \frac{1}{p}|\eta + tv|^p. \end{align*} Because $F$ is differentiable on $\mathbb{R}^n$ and $\gamma$ is differentiable with derivative $\gamma'(t)=v$, the [Chain Rule for Maps Between Euclidean Spaces](/theorems/323) gives \begin{align*} \varphi'(t) = \nabla F(\gamma(t))\cdot \gamma'(t) = A(\gamma(t))\cdot v. \end{align*} Evaluating this identity at $t=1$ and $t=0$ gives \begin{align*} \varphi'(1)-\varphi'(0) = A(\xi)\cdot v - A(\eta)\cdot v. \end{align*} Since $v=\xi-\eta$, this is exactly \begin{align*} \varphi'(1)-\varphi'(0) = \big(A(\xi)-A(\eta)\big)\cdot(\xi-\eta). \end{align*} Therefore it remains to prove that $\varphi'$ is nondecreasing on $[0,1]$, and strictly increasing when $v \neq 0$. [/guided] [/step] [step:Show the derivative along the segment is strictly increasing unless the segment is constant] Assume first that $v \neq 0$. Let $\mathcal{L}^1$ denote one-dimensional [Lebesgue measure](/page/Lebesgue%20Measure) on $\mathbb{R}$. On every open subinterval on which $\gamma(t) \neq 0$, the function $\varphi$ is twice differentiable and \begin{align*} \varphi''(t) = |\gamma(t)|^{p-2}|v|^2 + (p-2)|\gamma(t)|^{p-4}(\gamma(t)\cdot v)^2. \end{align*} For $\gamma(t)\neq 0$, the [Cauchy-Schwarz inequality](/theorems/432) applied to the Euclidean [inner product](/page/Inner%20Product) on $\mathbb{R}^n$ gives \begin{align*} (\gamma(t)\cdot v)^2 \leq |\gamma(t)|^2 |v|^2. \end{align*} If $p \geq 2$, both terms in the formula for $\varphi''(t)$ give $\varphi''(t)\geq |\gamma(t)|^{p-2}|v|^2>0$. If $1<p<2$, then \begin{align*} \varphi''(t) \geq |\gamma(t)|^{p-2}|v|^2 - (2-p)|\gamma(t)|^{p-2}|v|^2 = (p-1)|\gamma(t)|^{p-2}|v|^2 > 0. \end{align*} The equation $\gamma(t)=0$ has at most one solution because $v\neq 0$. If $t_0 \in (0,1)$ is such a solution, then $\gamma(r)=(r-t_0)v$, so near $t_0$ the possible singular factor is bounded by a constant multiple of $|r-t_0|^{p-2}$; this function is $\mathcal{L}^1$-integrable on compact intervals because $p-2>-1$. Hence $\varphi''$ is locally integrable on $(0,1)$ and satisfies $\varphi''(t)>0$ for all $t \in (0,1)$ except possibly one point. We now justify the integration of $\varphi''$ across the possible zero of $\gamma$. On every closed subinterval avoiding $t_0$, the function $\varphi'$ is continuously differentiable, so the [Fundamental Theorem of Calculus](/theorems/632) applies there. If $0 \leq s < t \leq 1$ and no such $t_0$ lies in $(s,t)$, this gives \begin{align*} \varphi'(t)-\varphi'(s) = \int_s^t \varphi''(r)\, d\mathcal{L}^1(r) > 0. \end{align*} If $t_0 \in (s,t)$, apply the same theorem on $[s,t_0-\varepsilon]$ and $[t_0+\varepsilon,t]$, where $0<\varepsilon<\min\{t_0-s,t-t_0\}$. Since $\varphi'$ is continuous on $[0,1]$ and $\varphi''$ is $\mathcal{L}^1$-integrable near $t_0$, passing to the limit as $\varepsilon \downarrow 0$ gives \begin{align*} \varphi'(t)-\varphi'(s) = \int_s^t \varphi''(r)\, d\mathcal{L}^1(r). \end{align*} The integrand is positive except possibly at $t_0$, so the integral over the interval of positive $\mathcal{L}^1$-measure is strictly positive. Thus $\varphi'$ is strictly increasing on $[0,1]$ when $v\neq 0$. If $v=0$, then $\gamma$ is constant and $\varphi'(t)=0$ for every $t \in [0,1]$. [/step] [step:Conclude monotonicity and characterize equality] Using the identity from the line-segment reduction, \begin{align*} \big(A(\xi)-A(\eta)\big)\cdot(\xi-\eta) = \varphi'(1)-\varphi'(0). \end{align*} If $\xi\neq \eta$, then $v\neq 0$, so $\varphi'$ is strictly increasing and therefore \begin{align*} \varphi'(1)-\varphi'(0) > 0. \end{align*} If $\xi=\eta$, then $v=0$, and directly \begin{align*} \big(A(\xi)-A(\eta)\big)\cdot(\xi-\eta)=0. \end{align*} Substituting $A(z)=|z|^{p-2}z$ gives \begin{align*} \big(|\xi|^{p-2}\xi - |\eta|^{p-2}\eta\big)\cdot(\xi-\eta) \geq 0, \end{align*} with equality if and only if $\xi=\eta$. This proves the theorem. [/step]