[proofplan] Verify the three norm axioms from the inner product axioms. Positivity and homogeneity are direct. The triangle inequality follows from Cauchy-Schwarz. [/proofplan] [step:Verify positivity] For any $x\in V$, the inner product axiom gives $(x,x)_V\ge 0$, so $\|x\|_V$ is defined and nonnegative. If $\|x\|_V=0$, then $(x,x)_V=0$, hence $x=0$ by positive definiteness. Conversely $\|0\|_V=0$. [/step] [step:Verify homogeneity] For any scalar $\lambda$ and any $x\in V$, \begin{align*} \|\lambda x\|_V^2 &=(\lambda x,\lambda x)_V \\ &=|\lambda|^2(x,x)_V \\ &=|\lambda|^2\|x\|_V^2. \end{align*} Taking square roots gives $\|\lambda x\|_V=|\lambda|\|x\|_V$. [/step] [step:Prove the triangle inequality] For $x,y\in V$, the [Cauchy-Schwarz Inequality](/theorems/432) for inner product spaces gives the bound $|(x,y)_V|\le \|x\|_V\|y\|_V$. Here $\operatorname{Re} z$ denotes the real part of a scalar $z$. Therefore \begin{align*} \|x+y\|_V^2 &=(x+y,x+y)_V \\ &=\|x\|_V^2+2\operatorname{Re}(x,y)_V+\|y\|_V^2 \\ &\le \|x\|_V^2+2 |(x,y)_V|+\|y\|_V^2 \\ &\le \|x\|_V^2+2\|x\|_V\|y\|_V+\|y\|_V^2 \\ &=(\|x\|_V+\|y\|_V)^2. \end{align*} Both sides are nonnegative, so $\|x+y\|_V\le \|x\|_V+\|y\|_V$. [/step] [step:Conclude the proof] The function $\|\cdot\|_V$ satisfies positivity, definiteness, homogeneity, and the triangle inequality, so it is a norm on $V$. [/step]