Cube of n minus 23 Greater than Square of (4n-7)

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Theorem

Let $n \in \Z$ such that $n \ge 12$.

Then:

$n^3 - 23 > \paren {4 n - 7}^2$


Proof

The proof proceeds by induction.

For all $n \in \Z$ such that $n \ge 12$, let $\map P n$ be the proposition:

$n^3 - 23 > \paren {4 n - 7}^2$


Basis for the Induction

$\map P {12}$ is the case:

\(\ds 12^3 - 23\) \(=\) \(\ds 1728 - 23\)
\(\ds \) \(=\) \(\ds 1705\)
\(\ds \paren {4 \times 12 - 7}^2\) \(=\) \(\ds 41^2\)
\(\ds \) \(=\) \(\ds 1681\)
\(\ds \leadsto \ \ \) \(\ds 12^3 - 23\) \(>\) \(\ds \paren {4 n - 7}^2\)

Thus $\map P {12}$ is seen to hold.


This is the basis for the induction.


Induction Hypothesis

Now it needs to be shown that if $\map P k$ is true, where $k \ge 12$, then it logically follows that $\map P {k + 1}$ is true.


So this is the induction hypothesis:

$k^3 - 23 > \paren {4 k - 7}^2$


from which it is to be shown that:

$\paren {k + 1}^3 - 23 > \paren {4 \paren {k + 1} - 7}^2$


Induction Step

This is the induction step:

\(\ds \paren {k + 1}^3 - 23\) \(=\) \(\ds k^3 + 3 k^2 + 3 k + 1 - 23\)
\(\text {(1)}: \quad\) \(\ds \) \(=\) \(\ds \paren {k^3 - 23} + \paren {3 k^2 + 3 k + 1}\)
\(\ds \paren {4 \paren {k + 1} - 7}^2\) \(=\) \(\ds \paren {\paren {4 k - 7} + 4}^2\)
\(\ds \) \(=\) \(\ds \paren {4 k - 7}^2 + 2 \times 4 \times \paren {4 k - 7} + 4^2\) Square of Sum
\(\text {(2)}: \quad\) \(\ds \) \(=\) \(\ds \paren {4 k - 7}^2 + 32 k - 40\) simplifying


Then we calculate:

\(\ds \paren {3 k^2 + 3 k + 1} - \paren {32 k - 40}\) \(=\) \(\ds 3 k^2 - 29 k + 40\)
\(\ds \) \(\ge\) \(\ds \paren {3 \times 12} k - 29 k + 40\) as $k \ge 12$
\(\ds \) \(=\) \(\ds 7 k + 40\) simplification
\(\ds \) \(>\) \(\ds 0\)
\(\text {(3)}: \quad\) \(\ds \leadsto \ \ \) \(\ds \paren {3 k^2 + 3 k + 1}\) \(>\) \(\ds \paren {32 k - 40}\)

Thus we have:

\(\ds \paren {k^3 - 23}\) \(>\) \(\ds \paren {4 k - 7}^2\) Induction Hypothesis
\(\ds \paren {3 k^2 + 3 k + 1}\) \(>\) \(\ds \paren {32 k - 40}\) from $(3)$
\(\ds \leadsto \ \ \) \(\ds \paren {k^3 - 23} + \paren {3 k^2 + 3 k + 1}\) \(>\) \(\ds \paren {4 k - 7}^2 + \paren {32 k - 40}\)
\(\ds \leadsto \ \ \) \(\ds \paren {k + 1}^3 - 23\) \(>\) \(\ds \paren {4 \paren {k + 1} - 7}^2\) from $(1)$ and $(2)$

So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.


Therefore:

$\forall n \in \Z_{\ge 12}: n^3 - 23 > \paren {4 n - 7}^2$

$\blacksquare$


Sources

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