If $\sigma(t) = (x(t),y(t))$ is a continuous curve in $\mathbb{R}^2$, I'd like to show that $\lVert \int_{a}^{b} \sigma(t) dt\rVert \leq \int_{a}^{b}\lVert\sigma(t) \rVert dt$ for all endpoints $a<b$.
I tried using Cauchy-Shwartz but it doesn't work and I don't know how to continue. I'm not sure if the inequality is true, even. Thanks!
Let $\sigma(t)=\big(x(t),y(t)\big)$. Then, setting $t_{n,k}=a+\frac{k(b-a)}{n}$, we have that $$ \int_a^b \sigma(t)\,dt=\left(\int_a^bx(t)\,dt,\int_a^by(t)\,dt\right)=\lim_{n\to\infty}\frac{b-a}{n}\sum_{k=1}^n\left(x(t_{n,k}),y(t_{n,k})\right), $$ and $$ \left\|\int_a^b \sigma(t)\,dt\,\right\|^2=\left(\int_a^bx(t)\,dt\right)^2+\left(\int_a^by(t)\,dt\right)^2 \\ =\left(\lim_{n\to\infty}\frac{b-a}{n}\sum_{k=1}^nx(t_{n,k})\right)^2+ \left(\lim_{n\to\infty}\frac{b-a}{n}\sum_{k=1}^n y(t_{n,k})\right)^2 \\=\lim_{n\to\infty}\frac{(b-a)^2}{n^2}\left(\left(\sum_{k=1}^n x(t_{n,k})\right)^2+\left(\sum_{k=1}^n y(t_{n,k})\right)^2\right)\\ \le \lim_{n\to\infty}\frac{(b-a)^2}{n^2}\left(\sum_{k=1}^n\big(x^2(t_{n,k})+y^2(t_{n,k})\big)^{1/2}\right)^2=\left(\int_a^b\big(x^2(t)+y^2(t)\big)^{1/2}\,dt\right)^2=\left(\int_a^b \|\sigma(t)\|\,dt\right)^2 $$ We have used above the Minkowski inequality.