Jensen's Implies Cauchy-Schwarz

We derive Cauchy-Schwarz via Jensen's inequality.

11 September 2025

In this, we aim to prove the following theorem

Theorem [Cauchy-Schwarz]. Given ui,vi∈Ru_i,v_i\in \R,

∑i=1nuivi≤∑i=1nui2∑i=1nvi2.\begin{equation} \sum_{i=1}^n u_iv_i\leq \sqrt{\sum_{i=1}^n u_i^2}\sqrt{\sum_{i=1}^n v_i^2}. \end{equation}

via the following inequality

Theorem [Jensen's]. Given a random vector Z∈RnZ\in \R^n and ff concave,

E[f(Z)]≤f(E[Z]).\begin{equation} \mathbb{E}[f(Z)]\leq f(\mathbb{E}[Z]). \end{equation}

The key is concavity of the function f(x,y):=xpy1−pf(x,y):=x^py^{1-p} for x,y>0x,y> 0 and 0≤p≤10\leq p\leq 1. Applying Jensen's to ff yields

E[f(X,Y)]≤f(E[X,Y]).\begin{equation} \mathbb{E}[f(X,Y)]\leq f(\mathbb{E}[X,Y]). \end{equation}

Let X,YX,Y be r.v.'s taking values in {xi},{yi}⊂R>0\{x_i\},\{y_i\} \subset \mathbb{R}^{>0}, both with probabilities {λi}\{\lambda_i\}. We analyzie the left and right hand sides of (4)(4).

LHS=E[XpY1−p]=∑i=1nλixipyi1−p\begin{equation} \begin{split} \text{LHS} &= \mathbb{E}[X^pY^{1-p}]\\ &=\sum_{i=1}^n \lambda_i x_i^py_i^{1-p} \end{split} \end{equation}

and

RHS=f(∑i=1nλixi,∑i=1nλiyi)=(∑i=1nλixi)p(∑i=1nλiyi)1−p.\begin{equation} \begin{split} \text{RHS} &= f\left(\sum_{i=1}^n \lambda_ix_i, \sum_{i=1}^n\lambda_iy_i\right)\\ &=\left(\sum_{i=1}^n \lambda_ix_i\right)^p\left(\sum_{i=1}^n \lambda_i y_i\right)^{1-p}. \end{split} \end{equation}

Now set λi=1/n\lambda_i=1/n and p=1/2p=1/2. Thus,

LHS=1n∑i=1nx1/2y1/2=1n∑i=1nxiyi\begin{equation} \begin{split} \text{LHS} &= \frac{1}{n}\sum_{i=1}^n x^{1/2}y^{1/2}\\ &=\frac{1}{n}\sum_{i=1}^n \sqrt{x_iy_i} \end{split} \end{equation}

and

RHS=∑i=1nxin∑i=1nyin=(1n)2∑i=1nxi∑i=1nyi.\begin{equation} \begin{split} \text{RHS} &=\sqrt{\sum_{i=1}^n \frac{x_i}{n}}\sqrt{\sum_{i=1}^n \frac{y_i}{n}} \\ &=\left(\sqrt{\frac{1}{n}}\right)^2 \sqrt{\sum_{i=1}^n x_i} \sqrt{\sum_{i=1}^n y_i}. \end{split} \end{equation}

Cancelling the 1/n1/n on both sides, and performing a change of variables ui:=xiu_i:=\sqrt{x_i} and vi:=yiv_i:=\sqrt{y_i} (recall x,y>0 so we can take square roots) yield

LHS=∑i=1nui2vi2=∑i=1nuivi\begin{equation} \begin{split} \text{LHS} &= \sum_{i=1}^n \sqrt{u_i^2v_i^2}\\ &=\sum_{i=1}^nu_iv_i \end{split} \end{equation}

and

RHS=∑i=1nui2∑i=1nvi2,\begin{equation} \begin{split} \text{RHS} &= \sqrt{\sum_{i=1}^n u_i^2} \sqrt{\sum_{i=1}^n v_i^2}, \end{split} \end{equation}

yielding the conclusion for ui,vi>0u_i,v_i>0. The other cases are obvious.