Proof $\sum\limits_{r=1}^{n} r >\frac{1}{2}n^2$ 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?
 A: Your second step should read $$\sum_{r=1}^k r > \dfrac{k^2}{2}$$
Then note that $$\sum_{r=1}^{k+1} r =  \underbrace{(k+1) + \sum_{r=1}^{k} r > (k+1) + \dfrac{k^2}{2}}_{\text{Induction hypothesis}} = \underbrace{\dfrac{k^2+2k+2}{2} > \dfrac{k^2+2k+1}{2}}_{a+\frac12 > a} = \dfrac{(k+1)^2}{2}$$
A: Hint $\ $ It is easy to prove by induction that an increasing function remains $\ge$ its initial value, i.e. that  $\rm\:f(n\!+\!1) \ge f(n)\:$ for $\rm\:n\ge 1\:$ $\Rightarrow$ $\rm\:f(n) \ge f(1)\:$ for $\rm n\ge 1$.
For $\rm\:f(n) = $ RHS - LHS of your inequality,  $\rm\:f\:$ is increasing since
$$\rm f(n+1) - f(n)\: =\ n\!+\!1 - ((n+1)^2 -n^2)/2 \: =\: 1/2\, >\, 0$$
So for all $\rm\:n\ge 1,\: f(n) \ge f(1) = 1/2,\:$ i.e. RHS $-$ LHS $\ge 1/2,\:$ so $\,$ RHS > LHS. $\quad$ QED
Note how this viewpoint removes all the guesswork from the inductive proof, reducing it to a simple rote algebraic calculation, viz. proving the positivity needed to show that $\rm\:f\:$ is increasing. The same technique works for many inductive problems of this type. For many examples see my prior posts on telescopy, where I show how the above can be viewed as the trivial induction that a sum of positive terms is positive.
A: \begin{align}
\sum_{1 \leq r \leq k+1}r &> \frac{(k+1)^2}{2}\\
\sum_{1 \leq r \leq k}r+(k+1) &> \frac{k^2}{2}+\frac{2k+1}{2}\\
2\sum_{1 \leq r \leq k}r+2(k+1) &> k^2+(2k+1)\\
2\sum_{1 \leq r \leq k}r &> k^2\\
 2(k+1) &> (2k+1)\\
\therefore \sum_{1 \leq r \leq k+1}r &> \frac{(k+1)^2}{2}
\end{align}
