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 :)

Denote $p(n):=(1+\frac{1}{1^3})(1+\frac{1}{2^3})(1+\frac{1}{3^3})...(1+\frac{1}{n^3})$. 

Claim:
$$p(n)\leq3-\frac2{n^2},\,\forall n\geq2.$$

For $n=2$, we have $\frac94\leq3-\frac12$.
Then suppose $p(n)\leq3-\frac2{n^2}$ for some $n$. We see
$$\eqalign{
p(n+1)&=p(n)(1+\frac1{(n+1)^3})\cr
&\leq3-\frac{2}{n^2}+\frac3{(n+1)^3}-\frac{2}{n^2(n+1)^3}\cr
&=3+\frac{3n^2-2(n^3+3n^2+3n+1)-2}{n^2(n+1)^3}\cr
&=3-\frac{2n^3+3n^2+6n+4}{n^2(n+1)^3}\cr
&=3-\frac{2n^3+2n^2+(n^2+6n+4)}{n^2(n+1)^3}\cr
&\leq3-\frac{2n^2(n+1)}{n^2(n+1)^3}\cr
&\leq3-\frac2{(n+1)^2}}.$$

As pointed out by @saulspatz, one can prove that $p(n)\leq3-\frac1n,\forall n\geq1$ by the same method.

Hope this helps.
