Q. What is $\min_{f \in \Bbb D} \int_{0}^{1} (1+x^2) f^{2}(x) dx$ where $\Bbb D = \{f : [0,1] \to \Bbb R : f \; \text {continuous}, \int_{0}^{1} f(x) dx=1\}$ ?
Attempt : I suppose $f^{2}(x)=(f(x))^2$.
By Schwarz's inequality, $1=(\int_{0}^{1} f(x) dx)^2 \le (\int_{0}^{1}1^2 dx)(\int_{0}^{1}(f(x))^2 dx)=\int_{0}^{1}(f(x))^2 dx \;\;\;\;\;\;\;\;\;\;(1)$
Using $(1)$ , $\int_{0}^{1} (1+x^2)f^{2}(x) dx=\int_{0}^{1} (f(x))^2 dx+\int_{0}^{1}x^2 (f(x))^2 dx \ge 1+\frac {m^2}3$ where $m=\min \{f(x) : x \in [0,1]\}$ and by compactness of $[0,1]$, $f$ takes it's minimum.
Since at $m=0$ we have minimum value of the function $1+\frac {m^2}3$ as $1$.
Hence $\min_{f \in \Bbb D} \int_{0}^{1} (1+x^2) f^{2}(x) dx=1$.
Is this solution correct? Thanks.
I don't think it is correct. You have two inequalities, and it is not clear that they are optimal for the same $f$ (or that the second one make sense with the condition $\int_0^1f\,dx=1$).
What if you instead write $$ 1=\int_0^1 f(x)\,dx=\int_0^1\frac{1}{\sqrt{1+x^2}}\sqrt{1+x^2}f(x)\,dx $$ and then apply the Cauchy-Schwarz inequality?