# An inequality involving integrals and square root

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!

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.
• This sup is nothing else than $\int _a^b \sqrt{\sum f_j^2}$, so I don't see how you get the inequality. – MikeTeX Jan 1 at 10:58
• Denote by $b_j$ the integral of $f_j$. What is sup of $\sum a_j b_j$ over all unit vectors $a$?. – Kavi Rama Murthy Jan 1 at 11:56
• It's OK, but your "sup" is misleading and unnecessarily complicates the matter. Simply define a priori $b_j$ to be the integral of $f_j$, $B$ to be the vector $(b_1, \ldots, b_n)$, $F$ to be $(f_1(x), \ldots, f_n(x))$ and $A$ to be the vector $B/||B||_2$ to get the result with Cauchy-Shwartz, as you did: $||B||_2 = A\cdot B = \int A\cdot F \leq \int ||F||_2$. (note: I've upvoted for your nice answer anyway). – MikeTeX Jan 1 at 19:43
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