# A calculus of the absurd

##### 4.12.3 Adding or removing terms

It seems “obvious”4040 But be careful; so does the pigeonhole principle (Section 6.5) and applying that is far from obvious to say that if $$a \leqq b$$ then the validity of the inequality is unchanged by any operation which either makes $$a$$ smaller or $$b$$ bigger, for example, if we have $$a \leqq b$$, then both of these inequalities must also be true

\begin{align} & a \leqq b + 1 \\ & a - 1 \leqq b \end{align}

Or, more generally, for any $$x > 0$$, $$a \leqq b$$ implies that both

\begin{align} & a \leqq b + x \\ & a - x \leqq b \end{align}

This is a fundamental (read: very useful) principle for manipulating inequalities.

• Example 4.12.1 Show that $$n! \geqq \left (\frac {n}{2}\right )^{\frac {n}{2}}$$

As when proving identities (see Section 6.2), we can prove this by starting with one side of the inequality and showing that this is smaller/larger than the other half. Starting with $$n!$$ we can apply the definition, which gives the equality that

\begin{align} n! &= n \times (n-1) \times (n-2) \times ... \times 3 \times 2 \times 1 \\ \end{align}

We now want to make the inequality smaller, and not just in any old way; we specifically want to make it smaller and look like $$\left (\frac {n}{2}\right )^{\frac {n}{2}}$$. It really helps to consider the structure of $$\left (\frac {n}{2}\right )^{\frac {n}{2}}$$ here - i.e. it looks something a bit like $$\frac {n}{2} \times \frac {n}{2} \times ... \times \frac {n}{2}$$ (or it would if we could be sure that $$\frac {n}{2}$$ was an integer, which we can by rounding $$\frac {n}{2}$$ up - it’s fine to show that $$n!$$ is bigger than (or equal to) something which is in turn bigger than (or equal to) $$\frac {n}{2}$$, as this still proves the fact that we are after - which we can write as $$\left \lceil \frac {n}{2} \right \rceil$$). Let us try to remove some terms to “fit” $$n!$$ to $$\left (\frac {n}{2}\right )^{\frac {n}{2}}$$

\begin{align} n! &\geqq n \times (n - 1) \times ... \times \left (\left \lceil \frac {n}{2} \right \rceil \right ) \\ \end{align}

We can then make all the terms smaller, like this

\begin{align} n! &\geqq \left (\left \lceil \frac {n}{2} \right \rceil \right ) \times ... \times \left (\left \lceil \frac {n}{2} \right \rceil \right ) \\ &= \left (\left \lceil \frac {n}{2} \right \rceil \right ) ^{\left \lceil \frac {n}{2} \right \rceil } \\ &\geqq \left (\frac {n}{2}\right )^{\frac {n}{2}}. \end{align}