I have a multiple choice question as follows. Now I can certainly attempt by putting in nice natural values for $n$, but I wish to know how can we arrive at these inequalities constructively.
The proposition that is not true for $n\gt 1$ ($n\in\mathbb{N}$) is
$(1)$ $\sum_{k=1}^{k=n} 1/k^2\lt 2-1/n$
$(2)$ $\sum_{k=1}^{k=n} 1/\sqrt{k}\gt \sqrt{n}$
$(3)$ $\sum_{k=1}^{k=n}1/(n+k)\gt 13/24$
$(4)$ $\prod_{k=1}^{k=n}(1-1/2k)\gt 1/\sqrt{3n+1}$
I can arrive at $(2)$ by rationalizing the inequality $\sqrt{r}+\sqrt{r+1}\gt 2\sqrt{r}$. But I'm not sure how we would arrive at the other ones.
A hint would be appreciated. Thanks
As tomi commented, if it's a multiple choice test and you don't have the time to prove everything properly, just plug $n = 2$ into each left hand side and see what comes out. In this case, assuming the problem is correctly posed (i.e. exactly one of those inequalities does not hold for all $n \geqslant 2$), that already reveals the correct answer. If $n = 2$ isn't sufficient to decide the matter, trying $n = 3$ next is a good idea. But if that also doesn't yield the answer, stepping back and thinking more about it soon becomes quicker than plugging in more $n$.
A good first approach thinking about it is eyeballing the behaviour of the left (and when appropriate also the right) hand sides. This of course works better the more you know.
First inequality
If you know that $$\sum_{k = 1}^{\infty} \frac{1}{k^2} = \frac{\pi^2}{6} \tag{1}$$ (this is pretty famous, thus many people know it long before they can prove it) and that $\pi^2 < 10$, then it's immediate that $$\sum_{k = 1}^n \frac{1}{k^2} < \frac{\pi^2}{6} < \frac{10}{6} = \frac{5}{3} = 2 - \frac{1}{3},$$ so the first inequality cannot fail for $n \geqslant 3$, whence checking $n = 2$ suffices. Even if you don't know $(1)$ it is not difficult to prove $$\sum_{k = 1}^{\infty} \frac{1}{k^2} < \frac{5}{3}, \tag{2}$$ for example by using telescoping: $$\sum_{k = 1}^{\infty} \frac{1}{k^2} = 1 + \sum_{k = 2}^{\infty} \frac{1}{k^2} < 1 + \sum_{k = 2}^{\infty} \frac{1}{k^2 - \frac{1}{4}} = 1 + \sum_{k = 2}^{\infty} \biggl(\frac{1}{k - \frac{1}{2}} - \frac{1}{k + \frac{1}{2}}\biggr) = 1 + \frac{1}{2 - \frac{1}{2}} = \frac{5}{3}.$$ But if you don't know $(1)$, then it's unlikely that you know $(2)$, and coming up with the idea that $(2)$ might a) be true and b) be useful, plus finding a proof of $(2)$ would be a waste of time, particularly since success isn't guaranteed.
If you don't know the behaviour of $\sum_{k = 1}^n 1/k^2$, so you can't immediately see that the first inequality holds for all $n$ except possibly a few small $n$, there are several ways to prove the inequality quickly. One way was given in tomi's answer, another is induction. That is always a good strategy for problems of this type. If you don't immediately have a better idea, look whether you see a quick proof by induction.
Here that works well, going from $n$ to $n+1$ the left hand side increases by $1/(n+1)^2$ and the right hand side increases by $$\Bigl(2 - \frac{1}{n+1}\Bigr) - \Bigl(2 - \frac{1}{n}\Bigr) = \frac{1}{n} - \frac{1}{n+1} = \frac{1}{n(n+1)} > \frac{1}{(n+1)^2},$$ so if the inequality holds for some $n$ it a fortiori also holds for all larger $n$.
Another quick way to prove the first inequality is available if $$\sum_{k = 2}^{\infty} \frac{1}{k(k-1)} = 1$$ has been proved previously via telescoping. Then, if one remembrs, one can see $$\sum_{k = 1}^{n} \frac{1}{k^2} = 1 + \sum_{k = 2}^{n} \frac{1}{k^2} < 1 + \sum_{k = 2}^{n} \frac{1}{k(k-1)} = 1 + \Bigl(1 - \frac{1}{n}\Bigr)$$ without computation. (The induction proof — this is just another way to write the previous — has been delegated to memory.)
Second inequality
For the second inequality one might see $\sum_{k = 1}^n 1/\sqrt{k} \approx 2\sqrt{n}$ at a glance, which shows that the inequality cannot fail for large $n$. If one knows the behaviour of the sum well enough one can even see that it holds for all $n > 1$. But few people know the behaviour of the sum that well at a point where they face this problem, so realistically one would convince oneself that the inequality holds for all $n > 1$ by eyeballing plus plugging in $n = 2$ and maybe $n = 3$ if one uses this strategy.
Induction works quite well here too. The left hand side increases by $1/\sqrt{n+1}$ and the right hand side by $$\sqrt{n+1} - \sqrt{n} = \frac{1}{\sqrt{n+1} + \sqrt{n}} < \frac{1}{\sqrt{n+1}},$$ so from equality for $n = 1$ or the validity of the inequality for $n = 2$ one obtains the validity of the inequality for all $n > 1$.
Another quick way uses comparison with an integral. From $$\sum_{k = 1}^n \frac{1}{\sqrt{k}} > \int_1^{n+1} \frac{dt}{\sqrt{t}} = 2\sqrt{n+1} - 2$$ it immediately follows that the inequality holds for $n \geqslant 3$ (because then $\sqrt{n+1} \geqslant 2$), and upon closer inspection one finds $2\sqrt{3} - 2 \approx 1.4641 > \sqrt{2}$, so this settles it completely. (But of one doesn't know the values of $\sqrt{3}$ and $\sqrt{2}$ well enough off the top of one's head it is quicker to settle the case $n = 2$ by plugging in, $1 + 1/\sqrt{2} > 1 + 1/2 > \sqrt{2}$ is quickly seen without knowing $\sqrt{2}$ well.)
Third inequality
If one is sufficiently familiar with logarithms and/or Riemann sums one sees $$\lim_{n \to \infty} \frac{1}{n+k} = \log 2 \approx 0.693147 > \frac{2}{3} = \frac{16}{24} > \frac{13}{24} = 0.541\overline{6}$$ and thus the inequality certainly holds for large enough $n$. Since there's a lot of space between the right hand side and the limit of the left "large enough" can actually be pretty small, but seeing that $n \geqslant 2$ is large enough without any calculation requires a lot of familiarity. But it's easy to see that $$\sum_{k = 1}^n \frac{1}{n+k} < \int_n^{2n} \frac{dt}{t} = \log 2$$ (since $1/t$ is decreasing), thus one may guess that the sum is increasing. If that guess is correct (it is, as we shall see) it suffices to calculate $$\sum_{k = 1}^{2} \frac{1}{2+k} = \frac{1}{3} + \frac{1}{4} = \frac{7}{12} > \frac{13}{24}.$$ To see that the sum increases with $n$, note $$\sum_{k = 1}^{n+1} \frac{1}{n+1+k} - \sum_{k = 1}^{n} \frac{1}{n+k} = \sum_{m = n+2}^{2m+2} \frac{1}{m} - \sum_{m = n+1}^{2n} \frac{1}{m} = \frac{1}{2n+1} + \frac{1}{2n+2} - \frac{1}{n+1} = \frac{1}{2n+1} - \frac{1}{2n+2} > 0.$$ This is also reached without first eyeballing the behaviour for large $n$ by just looking at how the sum changes when $n$ is replaced with $n+1$, which is a natural thing to do when one doesn't know the limiting behaviour.
Fourth inequality
To see the behaviour of the left hand side as $n$ increases is more difficult here than for the first three. Fortunately(?) trying to prove the inequality by induction already fails at the base case: $$\prod_{k = 1}^{2} \Bigl(1 - \frac{1}{2k}\Bigr) = \frac{1}{2}\cdot \frac{3}{4} = \frac{3}{8} = \sqrt{\frac{9}{64}} = \sqrt{\frac{63}{64}}\cdot \frac{1}{7} < \frac{1}{\sqrt{3\cdot 2 + 1}}.$$ If it didn't already fail for $n = 2$, say if the right hand side were $1/\sqrt{3n+2}$, or to delay the failure further $1/\sqrt{3n+10}$ it would be harder to show that the inequality is violated for some $n \geqslant 2$ (in fact, for all sufficiently large $n$).
One neat but not entirely obvious way is to write $$\prod_{k = 1}^{n}\Bigl(1 - \frac{1}{2k}\Bigr) = \prod_{k = 1}^{n} \frac{2k-1}{2k} = \prod_{k = 1}^{n} \frac{(2k-1)(2k)}{(2k)^2} = \frac{1}{2^{2n}}\binom{2n}{n}$$ and then use Stirling's approximation to conclude $$\prod_{k = 1}^{n}\Bigl(1 - \frac{1}{2k}\Bigr) \sim \frac{1}{\sqrt{\pi n}},$$ so that $$\prod_{k = 1}^{n}\Bigl(1 -\frac{1}{2k}\Bigr) < \frac{1}{\sqrt{3n + c}}$$ holds for all large enough $n$, whatever value $c$ is chosen.
If you know another quick and elementary way to prove (or refute, in the fourth case) one of these inequalities and think it doesn't merit its own answer, you're welcome to add it.