| $\left[\begin{array}{l}\text { Cauchy's Inequality } \\ \text { [ Completing the square. Sums of squares }\end{array}\right]$ |
| :--- |

$a+b=1$. What is the maximum value of the quantity $a b$?
$a(1-a)=1 / 4-(1 / 2-a)^{2}$.

## Solution

Since $a+b=1$, then $b=1-a$, and thus

$a b=a(1-a)=1 / 4-(1 / 2-a)^{2} \leq 1 / 4$,

since the square of any expression is non-negative.

## Answer

$1 / 4$.