Prove an integral inequality with squared integrals

Question

Given $f, g$ integratable prove that $$\left(\int_0^1 f(t) \ \mathrm{d}t\right)^2 + \left(\int_0^1 g(t) \ \mathrm{d}t\right)^2 \leq \left(\int_0^1 \sqrt{f^2(t) + g^2(t)} \ \mathrm{d}t\right)^2$$ I think that this exercise could be solved by applying Holder inequality, $$\left(\int_a^b f^p(t) \ \mathrm{d}t\right)^{1/p} + \left(\int_a^b g^p(t) \ \mathrm{d}t\right)^{1/p} \geq \left(\int_a^b (f(t) + g(t))^p \ \mathrm{d}t\right)^{1/p}$$ But I don't see how exactly. If you have any ideas, please share.

Answer 1

Consider the curve $\gamma: [0, 1] \to \mathbb{R}^2$ parametrized as $\gamma(t)=(x(t), y(t))$ with $$x(t) = \int_0^t f(x)\,dx \quad \text{ and } \quad y(t)=\int_0^t g(x)\,dx$$

The arclength of $\gamma$ is given by $$l(\gamma) = \int_0^1\sqrt{f(t)^2 + g(t)^2}\,dt$$ while the length of the line segment joining $\gamma(0)$ and $\gamma(1)$ is just $$\sqrt{\left(\int_0^1 f(t)\,dt\right)^2 + \left(\int_0^1 f(t)\,dt\right)^2}$$

The inequality follows since the line segment is the shortest path between any two points in $\mathbb{R}^2$.