<s>[INST] 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
 [/INST] 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 $\sqrt{15/8}$.
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