\[ \text{Note that when subbing things away in }LHS>RHS\text{ proofs}\\ \text{whenever you have nothing on both sides of the equation,}\\ \text{you use }=\text{ if the line exactly equals the line above}\\ \text{but }>\text{ if the }\textbf{line above}\text{ is greater than the current line.} \]
\[ \text{It also occasionally involves some fudgework as a consequence of the restriction on }n.\\ \text{Here, we note that }n \geq 1\\ \text{so we treat }k \geq 1\text{ in the assumption.} \]
\[ \text{Assuming that }k \geq 1\text{ in }\boxed{k^2 > k - 5}\text{ we have}\\ \begin{align*}(k+1)^2 &= k^2+2k + 1\\ &> (k - 5) + 2k+1 \tag{assumption}\\ &= 3k - 4 \end{align*}\]
Note that what we wished to prove was \( (k+1)^2 > (k+1) - 5\), i.e. \((k+1)^2 > k-4\). Well right now we have \(3k-4\). Is it necessarily true that \(3k-4 > k-4\)?
The answer here is yes, because if we solve that inequality we obtain \(k > 0\). This is convenient, because we've already assumed that \(k \geq 1\) to begin with. So if \(k > 0\) implies \(3k-4 > k-4\), then it is obviously also true that \(k \geq 1\) implies \(3k-4 > k-4\).
\[ \text{Thus continuing we have}\\ \begin{align*} (k+1)^2 &> 3k-4\\ &> k-4\text{ since k>1} \end{align*}\\ \text{which is as was required.} \]
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