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Suppose $p:[0,1]\to \mathbb C$ is a curve where $p(t)=u(t)+iv(t)$ and $u,v$ are smooth functions of $t$. Why then is $$\left(\int_0^1 \langle \dot{p},\dot{p}\rangle^{1\over 2} dt\right)^2\le \int_0^1 \langle \dot{p},\dot{p}\rangle dt$$ for any inner product $\langle ,\rangle$ and equality is achieved iff $\langle \dot{p},\dot{p}\rangle=c$ for some $c$?

My first thought is Cauchy-Schwarz inequality, but I am not too sure how it applies.

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You could apply CS to $f(t)=\langle \dot{p},\dot{p}\rangle^{1/2},\, g(t)=1$. – anon Jun 5 '12 at 19:15

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