Proving Inequality for Natural Numbers n>2

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Homework Statement



Proof that for n>2 and n is a natural number it holds that

\prod_{k=1}^{n}\frac{k^{2}+2}{k^{2}+1}<3

and
\prod_{k=1}^{n}\frac{k^{2}+2}{k^{2}+1}<\frac{3n}{n+1}

Homework Equations





The Attempt at a Solution



My best approach was to split the product over the fraction and then to arrive at a statement that looks like

\prod_{k=2}^{n}k^{2}+2<\prod_{k=1}^{n}k^{2}+1

I then tried to prove by induction that this statement holds but that doesn't really work. The best result I got (for n+1) is then

(\prod_{k=2}^{n}k^{2}+2)<(\prod_{k=1}^{n}k^{2}+1)\cdot\frac{n^{2}+2n+2}{n^{2}+2n+3}

But I can't do anything usefuel with that...
 
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You could try writing the product as the exponential of a sum, and then bounding the sum by an integral.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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