I'll let someone else post the one-liner. Here's my solution:

Solution: Summation(1/k^2+1/k^3)>(summation 1/k^2) Summation(1/k^2+1/k^3)>(summation 1/k^2)^2 There exists an integer such that Summation(1/k^2+1/k^3)>=m×(summation 1/k^2)^2

Note that this implies that $\zeta(3) \ge \zeta(2)^2 - \zeta(2)$.

asbodke wrote: Let $n\ge m\ge 1$ be integers. Prove that \[\sum_{k=m}^n \left (\frac 1{k^2}+\frac 1{k^3}\right) \ge m\cdot \left(\sum_{k=m}^n \frac 1{k^2}\right)^2.\] Raymond Feng and Luke Robitaille $$\sum_{k=m}^n \left (\frac 1{k^2}+\frac 1{k^3}\right) =\sum_{k=m}^n\frac{k+1}{k^3}\geqslant\dfrac{\left(\sum\limits_{k=m}^n \dfrac 1{k^2}\right)^2}{\sum\limits_{k=m}^n \dfrac 1{k(k+1)}}=\dfrac{1}{\frac 1m-\frac 1{n+1}}\left(\sum_{k=m}^n \frac 1{k^2}\right)^2>m\left(\sum_{k=m}^n \frac 1{k^2}\right)^2.\blacksquare$$

By the Cauchy-Schwarz inequality: $$\left(\sum_{k=m}^n \left(\frac{1}{k^2} +\frac{1}{k^3}\right)\right)\left(\sum_{k=m}^n \frac{1}{k+k^2}\right)\geq \left(\sum_{k=m}^n \frac{1}{k^2}\right)^2$$But $$\sum_{k=m}^n \frac{1}{k+k^2} = \sum_{k=m}^n \left(\frac 1k - \frac{1}{k+1}\right)=\frac 1m - \frac 1n < \frac 1m$$so $$\sum_{k=m}^n \left(\frac{1}{k^2} +\frac{1}{k^3}\right) \geq \frac{\left(\sum_{k=m}^n \frac{1}{k^2}\right)^2}{\sum_{k=m}^n \frac{1}{k+k^2} } > \frac{\left(\sum_{k=m}^n \frac{1}{k^2}\right)^2}{\frac 1m}= m\left(\sum_{k=m}^n \frac{1}{k^2}\right)^2$$

Different approach. Obviously, it's true for $(n,n-1)$. Now let's prove that if it's true for $(n,m)$, then it's also true for $(n,m-1)$. It's equivalent to show $$ \Big( \frac{1}{m^2}+\ldots + \frac{1}{n^2}\Big)^2 + \frac{1}{(m-1)^2}\geq 2\cdot \frac{1}{m-1}\cdot \Big( \frac{1}{m^2}+\ldots + \frac{1}{n^2}\Big)$$which is direct AM-GM, lol.

Note that, for all $n \leq i \leq m,$ $\dfrac{1}{i^2}+\dfrac{1}{i^3}=\dfrac{(1/i^2)^2}{1/(i(i+1))},$ and so by Cauchy-Schwarz, $\sum_{i=m}^{n} (\dfrac{1}{i^2}+\dfrac{1}{i^3}) \geq \dfrac{\displaystyle (\sum_{i=n}^{m} \dfrac{1}{i^2})^2}{\displaystyle \sum_{i=n}^{m} \dfrac{1}{i(i+1)}},$ hence it suffices to prove that $\displaystyle \sum_{i=n}^{m} \dfrac{1}{i(i+1)} \leq \dfrac{1}{m},$ which follows since $\dfrac{1}{i(i+1)}=\dfrac{1}{i}-\dfrac{1}{i+1}$ for all $i$, hence the above sum equals $\dfrac{1}{m}-\dfrac{1}{n+1}<\dfrac{1}{m}$, as desired.

More of a proof of concept for calc bash, not really feasible in contest unless you spend 3+ hours... Let $$A = \sum_{k = m}^n \frac{1}{k^2}$$and $$B = \sum_{k = m}^n \frac{1}{k^3}$$. It suffices to show $mB \ge (mA)(mA-1)$. The rest of the solution focuses on bounding $A, B$. We have $\frac{3}{2k^2} + \frac{3}{2(k+1)^2} \le \int_{k-1/2}^{k+1} \frac{1}{x^2} dx + \int_k^{k+3/2} \frac{1}{x^2} dx$ (difference $\frac{9}{2k^2(k+1)^2(2k-1)(2k+3)}$) hence $A \le \frac{1}{2m^2}+\frac{1}{3} (\int_{m-1/2}^{n+1/2} \frac{1}{x^2} dx + \int_m^n \frac{1}{x^2} dx + \int_{m+1/2}^{n-1/2} \frac{1}{x^2} dx) + \frac{1}{2n^2} \le \frac{1}{2m^2}+\frac{12m^2-1}{12m^3-3m}-\frac{1}{n+1}$. We have $3 \int_{k-1}^{k+1} \frac{1}{k^3} \le \frac{1}{(k-1)^3} + \frac{4}{k^3} + \frac{1}{(k+1)^3}$ when $k \ge 2$ (difference $\frac{12k^2-4}{(k^3-k)^3}$) hence $B \ge \frac{5}{6m^3} + \frac{1}{6(m+1)^3} + \frac{1}{2}\int_m^{m+1} \frac{1}{x^3} dx + \int_{m+1}^{n-1} \frac{1}{x^3} dx + \frac{1}{2}\int_{n-1}^n \frac{1}{x^3} dx + \frac{1}{6(n-1)^3} + \frac{5}{6(n+1)^3}$ if we assume signed integrals, so $\int_m^n = -\int_n^m$. Now we have $(mA)(mA-1)-mB \le (\frac{12m^2-1}{12m^2-3}+\frac{1}{2m}-\frac{m}{n+1})(\frac{2}{12m^2-3}+\frac{1}{2m}-\frac{m}{n+1})-\frac{m}{2(m+1)^2}-\frac{5}{6m^2}-\frac{m}{6(m+1)^3}-\frac{2m+1}{4m(m+1)^2}+\frac{1}{2(n-1)^2}$ It is obvious that we can now take all terms involving $n$ out, so it suffices to show $(\frac{12m^2-1}{12m^2-3}+\frac{1}{2m})(\frac{2}{12m^2-3}+\frac{1}{2m})-\frac{m}{2(m+1)^2}-\frac{5}{6m^2}-\frac{m}{6(m+1)^3}-\frac{2m+1}{4m(m+1)^2} \le 0$. Expanding we get $$-\frac{48m^7+192m^6+224m^5-288m^4-423m^3-52m^2+78m+21}{36(m+1)^3(m-4m^3)^2} \le 0$$which is obviously true when $m \ge 2$. For $m = 1$ we can simply check $\zeta(2) < \frac{5}{3}$ thus $(mA)(mA-1) < \frac{10}{9} < 1+\frac{1}{2^3}$ hence we are done. Remark: The bounds $\int_{k-1/2}^{k+1/2} \frac{1}{x^2}-\frac{1}{5x^4}-\frac{1}{k^2} dx \ge 0$ when $k \ge 2$ and $B \ge \frac{1}{2m^3} + \int_m^n \frac{1}{x^3} dx + \frac{1}{2n^3}$ may also suffice to finish the problem, possibly with less computation.

Fix $m$ and let $a_n:=\sum_{k=m}^n \left (\frac 1{k^2}+\frac 1{k^3}\right )-m \left (\sum_{k=m}^n \frac 1{k^2}\right )^2$. Then we can find $$a_n-a_{n+1}=\frac 1{(n+1)^2} \left (-1-\frac{1}{n+1}+\frac{m}{(n+1)^2}+2m\sum_{k=m}^n \frac 1{k^2}\right )=:\frac 1{(n+1)^2} d_n.$$Now we can make a claim. Claim: There exists a $k$ such that $d_n$ is positive if $n\geq k$ and negative otherwise. Proof. It's clear that $d_{n+1}-d_n>0$ and so $d_n$ is monotonic. And also that $d_m<0$ if $m>1$ (this is the $m$ that we fixed). Now note that $$d_{2m-1}=-1-\frac{1}{2m}+\frac{1}{4m}+2m \sum_{k=m}^{2m-1}\frac 1{k^2}\geq -1-\frac 1{4m}+2m (1+1+\dots+1)^2 \left (\sum_{k=1}^{2m-1} k^2\right)^{-1}=-1-\frac 1{4m}+2m\cdot m^2\left (\frac{m(m+1)(2m+1)}6\right)^{-1}$$where we have use the Schwarz inequality. It's easy to verify the RHS of above is positive for $m\geq 1$ and so $d_{2m-1}>0$. Now the claim follows by monotonicity. //// Notice that the claim implies that $a_k>a_{k+1}>a_{k+2}>\dots$ and $a_k>a_{k-1}>\dots>a_m$. It's clear that $a_m>0$ and so if we prove that $\lim_{n\to \infty} a_n$ is non-negative then we'll be done. Indeed, $$\lim_{n\to \infty} a_n=\zeta(2)-\sum_{k=1}^{m-1} \frac 1{k^2}+\zeta(3)-\sum_{k=1}^{m-1}\frac 1{k^3}-m\left (\zeta(2)-\sum_{k=1}^{m-1} \frac 1{k^2}\right )^2=:c_m.$$Next with a bit of manipulation we will find that $$c_n-c_{n+1}=\left(\zeta(2)-\sum_{k=1}^n \frac 1{k^2}-\frac 1n\right )^2\geq 0.$$So, $c_1\geq c_2\geq \dots \geq c_m\geq \dots$. Thus everything boils down to showing $\lim_{n\to \infty} c_n$ is non-negative. To do so we first calculate $$\alpha:=\lim_{n\to \infty} \frac{\zeta(2)-\sum_{k=1}^{n-1} \frac 1{k^2}}{1/\sqrt n}=\lim_{n\to \infty} \frac{\zeta(2)-\sum_{k=1}^{n-1} \frac 1{k^2}-\zeta(2)+\sum_{k=1}^{n-2} \frac 1{k^2}}{n^{-1/2}-(n-1)^{-1/2}}=\frac{\frac 1{(n-1)^2}}{(n-1)^{-1/2}-n^{-1/2}}=0.$$Here we used the Stolz-Cesaro theorem. Finally we have $\lim_{n\to \infty} c_n=-\alpha^2=0$, as desired.

Claim: $\sum^{n}_{k=m+1}(\frac{1}{k^2} + \frac{1}{k^3}) \geq (m+1)(\sum^{n}_{k=m+1}\frac{1}{k^2})^2$ implies $\sum^{n}_{k=m}(\frac{1}{k^2} + \frac{1}{k^3}) \geq (m)(\sum^{n}_{k=m}\frac{1}{k^2})^2$ Proof: Let $d = \sum^{n}_{k=m}\frac{1}{k^2}$ $$\frac{1}{m^2} + \frac{1}{m^3} \geq m(d)^2 - (m+1)(d-\frac{1}{m^2})^2$$$$m^2 + m \geq m^5d^2 - m^4(m+1)d^2 + 2m^2d(m+1) - (m+1)$$$$m^2 + m \geq - m^4d^2 + 2m^2d(m+1) - (m+1)$$$$m^4d^2 - 2m^2d(m+1) + (m+1)^2 \geq 0$$$$(m^2d - (m+1))^2 \geq 0$$which is true, thus the claim is true. Now we can reduce this into showing $\frac{1}{n^2} + \frac{1}{n^3} \geq n\frac{1}{n^4}$ which is obvious.

By Titu, \begin{align*} \frac{\displaystyle \left(\sum_{k = m}^n \frac{1}{k^2}\right)^2} {\displaystyle \sum_{k = m}^n \left(\frac{1}{k^2}+\frac{1}{k^3}\right)} \leq \sum_{k = m}^n \frac{\displaystyle \left(\frac{1}{k^2}\right)^2} {\displaystyle \frac 1{k^2}+\frac 1{k^3}} = \sum_{k = m}^n \frac{1}{k^2 + k} = \sum_{k = m}^n \left(\frac 1k - \frac{1}{k + 1}\right) = \frac 1m - \frac{1}{n + 1} \leq \frac 1m, \end{align*}done.

Of course, we bash. I'm around 50 percent sure this works. Replace all $n$ with $n - 1$, this does not change the problem. We need to show \[\sum_{m \le k < n}\left( \frac1{k^2} + \frac1{k^3} \right) \ge m\biggl( \sum_{m \le k < n}\frac1{k^2} \biggr)^2.\]Observe from integral bounds that \[\frac{n - m}{mn} = \int_m^n\frac1{x^2}\,\mathrm dx < \sum_{m \le k < n}\frac1{k^2} < \int_{m - 1}^{n - 1}\frac1{x^2}\,\mathrm dx = \frac{n - m}{(m - 1)(n - 1)}.\]Similarly, \[\frac{n - m}2 \cdot \frac{m + n}{m^2n^2} < \sum_{m \le k < n}\frac1{k^3}.\] It's enough to show (by substituting and dividing through by $n - m$) \[\frac1{mn} + \frac{m + n}{2m^2n^2} \ge \frac{n - m}{(m - 1)^2(n - 1)^2}.\]Using the estimates $m + n \ge n$ and $n - m \le n$ give \[\frac1{mn} + \frac1{2m^2n} \ge \frac n{(m - 1)^2(n - 1)^2},\]which is equivalent to \[m + \frac12 \ge \frac{m^2n^2}{(m - 1)^2(n - 1)^2}. \qquad (\bigstar).\]A weaker form is \[m + \frac12 \ge \left( \frac m{m - 1} \right)^4\]which holds if $m \ge 4$. So we only need to check the cases $m = 1$, $m = 2$ and $m = 3$. The case $m = 1$ is trivial. If $m = 3$, then for $n \ge 6$, the inequality $(\bigstar)$ holds, so we only need to check $n = 4$ and $n = 5$. The case $n = 4$ works since $\frac19 + \frac1{27} \ge 3 \cdot (3^{-2})^2$, and the case $n = 5$ works since \[\frac19 + \frac1{16} + \frac1{27} + \frac1{64} \ge 4 \cdot \left( \frac19 + \frac1{16} \right)^2.\] So this leaves the case $m = 2$. This is by far the most precise. Since $2x^2 - x$ is increasing when $x > \frac12$, we bound \[\sum_{k = 2}^n\frac1{k^2} < \frac{\pi^2}6 < \frac16 \cdot \left( \frac{22}7 \right)^2 = \frac{95}{147}.\]A terrible, terrible computation gives \[\sum_{2 \le k < 7}\frac1{n^3} \ge 2 \cdot \left(\frac{95}{147}\right)^2 - \frac{95}{147}.\]Therefore, we only need to check the cases $m = 2$ and $n = 3,4,5,6$, which is easy since $\frac14 + \frac19 + \frac1{16} + \frac1{25} < \frac12$.