An inequality involving integrals and square root

Question

Could someone help me with this inequality?

$$\left( \int_a^b \sqrt{\sum_{j=1}^n f_j^2(x)} dx \right)^2 \geq \sum_{j=1}^n \left(\int_a^b f_j(x) dx \right)^2$$

I tried to used the concavity of square root and then Cauchy Schwarz, but the latter is in the wrong direction. I appreciate any help!

Answer 1

Let $(a_1,a_2,\cdots,a_n)$ be a unit vector in $\mathbb R^{n}$. Then $\sum_j a_j \int_a^{b} f_j (x) dx=\int_a^{b}\sum_j a_j f_j (x) dx \leq \int_a^{b} \sqrt {\sum f_j^{2}}$ by C-S inequlaity. By taking sup over all u nit vectors $(a_1,a_2,\cdots,a_n)$ we get the desire dinequality.

Answer 2

Hint. Show the inequality for $n=2$, then the induction step is easy: $$ \sum_{j=1}^{n+1}\left(\int_{a}^{b}f_{j}(x)dx\right)^{2} =\sum_{j=1}^{n}\left(\int_{a}^{b}f_{j}(x)dx\right)^{2}+\left(\int_{a}^{b}f_{n+1}(x)dx\right)^{2}\\ \leq \left(\int_{a}^{b} \sqrt{\sum_{j=1}^{n}f_{j}(x)^{2}}dx\right)^{2}+\left(\int_{a}^{b}f_{n+1}(x)dx\right)^{2}\leq \left(\int_{a}^{b} \sqrt{\sum_{j=1}^{n+1}f_{j}(x)^{2}}dx\right)^{2}.$$

Hint for $n=2$. Note that $$|f_{1}(x)|=\sqrt{\sqrt{f_{1}(x)^{2}+f_{2}(x)^{2}}+f_{2}(x)}\cdot \sqrt{\sqrt{f_{1}(x)^{2}+f_{2}(x)^{2}}-f_{2}(x)}$$ then apply the Cauchy–Schwarz inequality