Proof $sumlimits_{r=1}^{n} r >frac{1}{2}n^2$ using induction

Question:

$$\text{Prove by induction that, for all integers } n, n \geq 1:$$
$$\sum\limits_{r=1}^{n} r >\frac{1}{2}n^2$$

Working:

Step 1 (Prove true for n=1):

$$1>\frac{1}{2}(1)^2$$

Step 2 (Assume true for n=k):
$$ k >\frac{1}{2}k^2$$

Step 3 (Prove true for n=k+1):

And having only faced equations with an equals (=) sign I have no idea what to do next. Right now I have assumed that it stands true for $k$ and I will try to prove for $k+1$. What should be my next step?

Answer

Your second step should read $$\sum_{r=1}^k r > \dfrac{k^2}{2}$$

Then note that $$\sum_{r=1}^{k+1} r = \underbrace{(k+1) + \sum_{r=1}^{k} r > (k+1) + \dfrac{k^2}{2}}_{\text{Induction hypothesis}} = \underbrace{\dfrac{k^2+2k+2}{2} > \dfrac{k^2+2k+1}{2}}_{a+\frac12 > a} = \dfrac{(k+1)^2}{2}$$