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: 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.
A: The inequality in the proposed solution is derived from the Cauchy-Schwarz inequality. Assuming $f,g$ are square-integrable on $[a,b]$, then we have the inequality
$$\left( \int_a^b f(x) \cdot g(x)  \; dx \right)^2 \le \left( \int_a^b (f(x))^2 \;dx \right) \cdot \left( \int_a^b (g(x))^2 \;dx \right)$$
or perhaps more compactly
$$\left( \int_a^b f \cdot g \right)^2 \le \left( \int_a^b f^2 \right) \cdot \left( \int_a^b g^2 \right)$$
Square both sides of the inequality in your post to see this. More explicitly: take $f(x) = \sqrt{1+x}, g(x) = \sqrt{1+x^3}$, then take the (positive) square root of the resulting inequality to bring it into the form presented in the original post.

To address the point of "isn't this seemingly-unrelated thing Cauchy-Schwarz?" from the comments, we note that these really just are specific instances of the more general Cauchy-Schwarz inequality. Let $\langle \cdot , \cdot \rangle$ denote an inner product over an inner product space. Let $u,v$ be members of said space. Then the Cauchy-Schwarz inequality says
$$\vert \langle u,v \rangle \vert^2 \le \langle u,u \rangle \cdot \langle v,v \rangle$$
This more general formulation is better elaborated on Wikipedia.
In your specific case, the space is - if I remember correctly - the space of square-integrable real functions on the interval $[a,b]$. Members of this space are such functions, and the inner product is defined by
$$\langle f,g \rangle = \int_a^b f(x) \cdot g(x)  \; dx$$
A: Not an answer, just some thoughts.
In this case we can also try using the arithmetic mean - geometric mean inequality:
$$\int _{0}^1 \sqrt {(1+x)(1+x^3)} dx \leq \frac{1}{2} \int _{0}^1 (1+x) dx+\frac{1}{2} \int _{0}^1 (1+x^3) dx = \\ = \frac{3}{4}+\frac{5}{8}=\frac{11}{8}=1.375$$
Meanwhile:
$$\sqrt{\frac{15}{8}}=1.3693 \ldots < \frac{11}{8}$$
So AMGM inequality is not as good as Cauchy-Schwartz.
Among other famous means, we also have the logarithmic mean, which is between arithmetic and geometric means:
$$\sqrt{ab} \leq l(a,b)=\frac{a-b}{\log (a/b)} \leq \frac{a+b}{2}$$
So curiously, we have:
$$\int _{0}^1 \sqrt {(1+x)(1+x^3)} dx \leq \int_0^1 \frac{(x-x^3) dx}{\log \frac{1+x}{1+x^3}} \leq \frac{11}{8}$$
This last integral is clearly too complicated to evaluate exactly here, but numerically we have:
$$\int_0^1 \frac{(x-x^3) dx}{\log \frac{1+x}{1+x^3}}=1.3703117346 \ldots$$
Which is still worse estimate than $\sqrt{15/8}$. Such is the power of Cauchy-Schwartz.
As for the lower bound, the harmonic mean is a good way:
$$\int _{0}^1 \sqrt {(1+x)(1+x^3)} dx \geq 2 \int _{0}^1 \frac{(1+x)(1+x^3)}{2+x+x^3} dx= 3-\frac{12}{\sqrt{7}} \arctan \frac{1}{\sqrt{7}}=1.36099 \ldots$$
Which is a good lower bound, since now we know:
$$1.36099 \ldots \leq \int _{0}^1 \sqrt {(1+x)(1+x^3)} dx \leq 1.36930 \ldots$$
And finally, the exact value:
$$\int _{0}^1 \sqrt {(1+x)(1+x^3)} dx= \frac{3}{4}+ \frac{9}{16} \log 3=1.367969 \ldots$$
Which is very close to $\sqrt{15/8}$.
