# Proving Holder inequality using Lagrange multipliers

1. Let $x \geq 0, y \geq 0$ and $p > 0, q > 0$ with $\dfrac{1}{p} + \dfrac{1}{q} = 1$. Show that $$xy \leq \frac{1}{p}x^p + \frac{1}{q}y^q.$$
2. Given $v = (v_1,\cdots,v_m) \in \mathbb{R}^{m}$, define $\displaystyle |v|_{p} = \left(\sum_{i}|v_{i}|^p \right)^{\frac{1}{p}}$. Use the previous inequality for show that, if $\displaystyle \frac{1}{p} + \frac{1}{q} = 1$ so, given $u,v \in \mathbb{R}^{m}$ the Holder inequality $|\langle u,v \rangle| \leq |u|_{p}|v|_{q}$ is true.

My idea is: Show that $S = \lbrace (x,y)\in \mathbb{R}^{2}\mid x \geq 0, y \geq 0, xy=1 \rbrace$ is a surface of $\mathbb{R}^{3}$ and defining the function $\varphi: \mathbb{R}^{2}_{+} \to \mathbb{R}$ by $\varphi(x,y) = xy$, $S = \varphi^{-1}(1)$. Then I want to apply Lagrange Multipliers to determine the minimum of $f|_M$ where\begin{align*} f: &&\mathbb{R}^{2} &\longrightarrow \mathbb{R},\\ &&(x,y) &\longmapsto \frac{1}{p}x^p + \frac{1}{q}y^q \end{align*} with $p,q > 0$ and $\displaystyle \frac{1}{p} + \frac{1}{q} = 1$.

I know that probably is the way, or at least a part of it. But I cannot determine the critical points. Can someone help me?

## 1 Answer

$\def\vec{\boldsymbol}$For the first question, note that the inequality obviously holds if $x = 0$ or $y = 0$, thus it suffices to prove for $x, y > 0$. Next, under the condition $xy = 1$, if $x \geqslant p^{\frac{1}{p}}$ or $y \geqslant q^{\frac{1}{q}}$, then the inequality also holds. Therefore, it suffices to look for minimum points of$$f(x, y) = \frac{1}{p} x^{\frac{1}{p}} + \frac{1}{q} y^{\frac{1}{q}} - xy$$ in the bounded region $D = (0, p^{\frac{1}{p}}) × (0, q^{\frac{1}{q}})$ under the condition $xy = 1$. Define$$F(x, y; λ) = f(x, y) - λ(xy - 1).$$ If $f$ reaches its minimum in the region $D$ under the condition $xy = 1$ at $(x_0, y_0)$, then there exists $λ_0 \in \mathbb{R}$ such that$$\frac{\partial F}{\partial x}(x_0, y_0; λ_0) = \frac{\partial F}{\partial y}(x_0, y_0; λ_0) = \frac{\partial F}{\partial λ}(x_0, y_0; λ_0) = 0,$$ which implies$$\begin{cases} x_0^{p - 1} - λ_0 y_0 = 0\\ y_0^{q - 1} - λ_0 x_0 = 0\\ x_0 y_0 = 1 \end{cases} \Longrightarrow \begin{cases} x_0^p = y_0^q = λ_0\\ x_0 y_0 = 1 \end{cases} \Longrightarrow λ_0 = λ_0^{\frac{1}{p} + \frac{1}{q}} = x_0 y_0 = 1,$$ thus $(x_0, y_0) = (1, 1)$. Therefore,$$\inf_{\substack{(x, y) \in D\\xy = 1}} f(x, y) \geqslant \min\left( f(1, 1), \min_{(x, y) \in \partial D} f(x, y) \right) = 0,$$ i.e. $\dfrac{1}{p} x^{\frac{1}{p}} + \dfrac{1}{q} y^{\frac{1}{q}} \geqslant xy$.

For the second question, if $\vec{u} = \vec{0}$ or $\vec{v} = \vec{0}$, then the inequality obviously holds. For $\vec{u}, \vec{v} ≠ \vec{0}$, since $\|\vec{u}\|_p, \|\vec{v}\|_q > 0$, using the inequality in the first question,\begin{align*} &\mathrel{\phantom{=}}{} \frac{|\langle \vec{u}, \vec{v} \rangle|}{\|\vec{u}\|_p \|\vec{v}\|_q} = \left| \sum_{k = 1}^n \frac{u_k v_k}{\|\vec{u}\|_p \|\vec{v}\|_q} \right| \leqslant \sum_{k = 1}^n \frac{|u_k| · |v_k|}{\|\vec{u}\|_p \|\vec{v}\|_q} = \sum_{k = 1}^n \frac{|u_k|}{\|\vec{u}\|_p} · \frac{|v_k|}{\|\vec{v}\|_q}\\ &\leqslant \sum_{k = 1}^n \left( \frac{1}{p} · \frac{|u_k|^p}{\|\vec{u}\|_p^p} + \frac{1}{p} · \frac{|v_k|^q}{\|\vec{v}\|_q^q} \right) = \frac{1}{p} · \frac{1}{\|\vec{u}\|_p^p} \sum_{k = 1}^n |u_k|^p + \frac{1}{q} · \frac{1}{\|\vec{v}\|_q^q} \sum_{k = 1}^n |v_k|^q\\ &= \frac{1}{p} + \frac{1}{q} = 1, \end{align*} i.e. $|\langle \vec{u}, \vec{v}\rangle| \leqslant \|\vec{u}\|_p \|\vec{v}\|_q$.