<s>[INST] Inequality $\sum_{k=1}^n\frac{k^2}{n^3+n^2}\le\sum_{k=1}^n\frac{k^2}{n^3+k^2}\le\sum_{k=1}^n\frac{k^2}{n^3+1}$ From this question How do you calculate this limit? $\lim_{n\to\infty}( \frac{1^2}{n^3+1^2} + \frac{2^2}{n^3+2^2} ... + \frac{n^2}{n^3+n^2})=?$
I have not understood the reason of the @xpaul's answer for the $(1)$,
$$\sum_{k=1}^n\frac{k^2}{n^3+n^2}\le\sum_{k=1}^n\frac{k^2}{n^3+k^2}\le\sum_{k=1}^n\frac{k^2}{n^3+1} \tag 1$$
 [/INST] Since the sum ranges from $k=1$ to $k=n$, we have $k\geq 1$ and $k\leq n$; thus $n^2\geq k^2\geq 1$. Translate by $n^3$ to get $n^3+n^2\geq n^3+k^2\geq n^3+1$, then invert (changing the direction of the inequalities) to get $\dfrac1{n^3+n^2}\leq \dfrac1{n^3+k^2}\leq \dfrac1{n^3+1}$. Multiply by $k^2$ to get $\dfrac{k^2}{n^3+n^2}\leq\dfrac{k^2}{n^3+k^2}\leq\dfrac{k^2}{n^3+1}$, and then just sum termwise over $k$.
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