Show that $\left(1+\frac{1}{1^3}\right)\left(1+\frac{1}{2^3}\right)\left(1+\frac{1}{3^3}\right)\cdots\left(1+\frac{1}{n^3}\right) < 3$ I have this problem which says that for any positive integer $n$, $n \neq 0$ the following inequality is true: $$\left(1+\frac{1}{1^3}\right)\left(1+\frac{1}{2^3}\right)\left(1+\frac{1}{3^3}\right)\cdots\left(1+\frac{1}{n^3}\right) < 3$$
This problem was given to me in a lecture about induction but any kind of solution would be nice.And also I'm in 10th grade :)

By induction we can prove the stronger 
$$\prod_{k=1}^n\left(1+\frac{1}{k^3}\right)<3\left(1-\frac{1}{n}\right)$$
indeed
1. base cases: by inspection the inequality is satisfied for for
    $n=1,2, 3$
2. induction step: 


*

*assume true that (Ind. Hyp.): $\prod_{k=1}^n\left(1+\frac{1}{k^3}\right)<3\left(1-\frac{1}{n}\right)$

*we want to prove that: $\prod_{k=1}^{n+1}\left(1+\frac{1}{k^3}\right)<3\left(1-\frac{1}{n+1}\right)$
then we have
$$\prod_{k=1}^{n+1}\left(1+\frac{1}{k^3}\right)=\prod_{k=1}^n\left(1+\frac{1}{k^3}\right) \cdot \left(1+\frac{1}{(n+1)^3}\right)<$$
$$\stackrel{Ind. Hyp.}<3\left(1-\frac{1}{n}\right)\left(1+\frac{1}{(n+1)^3}\right)\stackrel{?}<3\left(1-\frac{1}{n+1}\right)$$
thus we need to show that
$$3\left(1-\frac{1}{n}\right)\left(1+\frac{1}{(n+1)^3}\right)\stackrel{?}<3\left(1-\frac{1}{n+1}\right)$$
which is true indeed
$$1+\frac{1}{{n+1}^3}-\frac1n-\frac{1}{n(n+1)^3}\stackrel{?}<1-\frac{1}{n+1}$$
$$n-(n+1)^3-1\stackrel{?}<-n(n+1)^2$$
$$n-n^3-3n^2-3n-1-1\stackrel{?}<-n^3-2n^2-n$$
$$n^2+n+2\stackrel{?}>0$$
