<s>[INST] Proofs involving summations I am working on the following proof. 
Prove that $\displaystyle \sum_{k=1}^{n} \frac{2k^4+1}{k+2} \geq \frac{n(n+1)(2n+1)}{6}$.
My first question is if the above is equivalent to saying "Prove that:$\displaystyle \sum_{k=1}^{n} \frac{2k^4+1}{k+2} \geq \displaystyle \sum_{n=1}^{t} \frac{n(n+1)(2n+1)}{6}$." If this was in fact correct, I examined the partial sums and found
$1+\frac{33}{4}+\frac{163}{5}+\frac{171}{2}+ \cdots \geq 1+2^2+3^2+4^2+ \cdots$. From here I see that when $n=1$ the inequality turns into an equality. My plan from here was to use some kind of induction to show that $\forall k_{i}, i,j \in \mathbb{N}_, k_{i} \geq n_{j}$.
Other than this method I do not see another way to go, could someone offer a hint of a different route or some help with my current route. 
For my current route I am confused on the bounds for the second summation $\displaystyle \sum_{n=1}^{t} \frac{n(n+1)(2n+1)}{6}$.
Thank you.
 [/INST] I think what you are asking is $\frac{2k^4 + 1}{k + 2} \ge \frac{k(k+1)(2k+1)}{6}$.
$\frac{k(k+1)(2k+1)}{6}= \frac{k(k+1)(k+2)(2k+1)}{6(k+2)}=\frac{\frac 16(2k^4 + 7k^3 +7k^2 + 2k)}{k+2}$
So your question is true if and only if $2k^4 + 1 \ge \frac 16(2k^4 + 7k^3 +7k^2 + 2k)$
Which is true if and only if $ 10k^4 + 6 \ge 7k^3 + 7k^2 + 2k$
Which is true if and only if $10k + 6/k^4 \ge 7 + 7/k  + 2/k^2$.
Which is true for $k = 1$ and as the LHS increases and the RHS decreases, this is always true.
Which is something I think the authors of the problem didn't notice. 
So $\sum_{k=1}^{n}\frac{2k^4 + 1}{k + 2} = \sum_{k=1}^{n-1}\frac{2k^4 + 1}{k + 2}+ \frac{2n^4 + 1}{n+2}$
$\ge 0 +  \frac{2n^4 + 1}{n+2} =  \frac{2n^4 + 1}{n+2}$
$\ge \frac{n(n+1)(2n+1)}{6}$
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