Prove that $\int _{0}^1 \sqrt {(1+x)(1+x^3)} ≤ \sqrt {15/8}$ 
Prove that
$$\int _{0}^1 \sqrt {(1+x)(1+x^3)}\ dx ≤ \sqrt {15/8}$$

Now in this question, the solution that was provided me was $\int _{0}^1 \sqrt {(1+x)(1+x^3)}dx ≤ \sqrt{\int _{0}^1(1+x)dx \int _{0}^1(1+x^3)dx}$
I have never even heard or seen this in inequality before. Can anyone explain me what they are doing or if it is a very standard result in mathematics what that result or theorem is. please help

A simple proof of Cauchy-Schwartz:
Let $f(x)$ and $g(x)$ be two real square integrable functions in $[a,b]$.
Then, for all real values of $t$,
$$\int_{a}^{b} (f(x)+t g(x))^2  \ge 0  \Rightarrow t^2\int_{a}^{b} g^2(x) +2t \int_{a}^{b} f(x)g(x) dx + \int_{a}^{b}f^2(x) dx \ge 0, \forall t \in R.$$
Then the condition of the positivity of the quadratic $B^2 \le 4AC$ gives
$$\left ( \int_{a}^{b} f(x) g(x) dx \right)^2 \le \int_{a}^{b} f^2(x) dx \int_{a}^{b} g^2(x) dx $$ This is the integral form of Cauchy-Schwartz inequality.
