Prove that $(1+ \frac{1}{1^3})(1+\frac{1}{2^3})...(1+\frac{1}{n^3})<3$ [duplicate]

Question

I have tried to use induction, but after I assume that P(n) is true, I can't go further to prove that P(n+1) is true as well. I also have tried to find an intermediate inequality, but I can't figure out which inequality I should start from.

Something that seemed to be useful was taking P(n) and multiplying it by $(1+\frac{1}{(n+1)^3})$, therefore I have come to this

$(1+ \frac{1}{1^3})(1+\frac{1}{2^3})...(1+\frac{1}{n^3})<3 | \times(1+\frac{1}{(n+1)^3})$

$(1+ \frac{1}{1^3})(1+\frac{1}{2^3})...(1+\frac{1}{n^3})(1+\frac{1}{(n+1)^3})<3(1+\frac{1}{(n+1)^3})$

but, as anyone could imagine, I came to contradiction because I've tried to prove that $3(1+\frac{1}{(n+1)^3})<3$ which is false.

Any help it would be useful.

Answer 1

Using the fact $1+x\le e^x$ for all real $x,$ we have $$\left(1+ \frac{1}{1^3}\right)\left(1+\frac{1}{2^3}\right)\cdots\left(1+\frac{1}{n^3}\right)\le \dfrac{9}{4}\exp\left(\sum_{k=3}^{n}\dfrac{1}{k^3}\right).$$ Now use the fact that $$\sum_{k=1}^{\infty}\dfrac{1}{k^3}\lt\dfrac{\pi^2}{7}.$$