For real numbers \(A\) and \(B\) the Arithmetic-Geometric Mean Inequality says:
\(\bullet\quad A\cdot B\leq\left(\frac{A+B}2\right)^2.\)
\(\bullet\quad\mathrm{~if~}A,B\geq0\mathrm{~then~}\sqrt{AB}\leq\frac{A+B}2.\)
The quantity \(\sqrt{AB}\) is called the geometric mean, and \(\frac{A+B}2\) is the arithmetic mean of \(A\) and \(B\). These inequalities become equalities if and only if \(A=B\).
We can use the AGM inequality to find maximum and minimum values by substituting for \(A\) and \(B\) above. Keep in mind that we have multiplication on the “smaller” side and addition on the “greater” side. Thus, we can find a maximum for the multiplication side \((A\cdot B\leq M)\) if the addition side can be dealt with and a minimum for the addition side \((M\leq\left(\frac{A+B}{2}\right)^2)\) if the multiplication side can be dealt with.
Example: Find the maximum area of a rectangle with a perimeter of \(20m\). What dimensions give the maximum area?
Solution: We would like to find a maximum for the area. The area of a rectangle is given by \(A=l\cdot w\), a maximum for the area would then mean finding a number \(\boldsymbol{M}\text{ so that }\boldsymbol{l}\cdot\boldsymbol{w}\leq\boldsymbol{M}\).
We’re told the perimeter is \(20m\), this is given by \(2l+2w=20,\mathrm{i.e.},l+w=10.\) We use AGM by setting \(A=l\operatorname{and}B=w\) which gives:
This is useful because the left side contains \(\boldsymbol{l}\cdot\boldsymbol{w}\) and on the right side we use \(l+w=10\) to look for our \(M\).
\[(l)(w)\leq\left(\frac{l+w}2\right)^2\\l\cdot w\leq\left(\frac{10}2\right)^2\\l\cdot w\leq25\]
Therefore, we see that the maximum area is \(25m^2\).
We want to know when \(l\cdot w=25\). AGM says this happens when \(A=B\), that is \(l=w\) which we can use with \(l+w=10\):
\[\begin{aligned}l+l&=10\\l&=5\end{aligned}\]
Substituting \(l=5\) into \(l = w\) tells us that \(w = 5\), meaning the rectangle with the maximum area is a \(5\leftrightarrow5\) square.
Example: Find a lower bound for \(\text{r }f(x)=x+\frac9x+1\) when x>0. Find the point(s) where this happens.
Solution: We’re being asked to find a lower bound, or a minimum, so we’ll have to use an inequality. The key thing to notice is that we have \(x+\frac9x\), if we multiply \(x\) and \(\frac9x\) we can eliminate the \(x\); this will allow us to use AGM. Since we have \(x>0\) we can use the second version with \(A=x\mathrm{~and~}B=\frac9x:\)
\[\sqrt{AB}\leq\frac{A+B}2\\\\(\sqrt{x\frac9x}\leq\frac{x+\frac9x}2\\2\sqrt{9}\leq x+\frac9x\\6\leq x+\frac9x\]
At this point we almost have the function. We can add 1 to both sides to complete it:
\[\begin{aligned}&7\leq x+\frac9x+1\\&7\leq f(x)\end{aligned}\]
We want to know \(x\) where \(f(x)=7\), according to AGM this happens when \(A=B\):
\[x=\frac9x\\x^2=9\\x=\pm3\]
Since we know that \(x>0\) the solution is \(x=3\). Thus, the point is (3,7).