a^{2} < a + b < ab < b^{2}
The prompt was devised by Mark Greenaway (an Advanced Skills Teacher in Suffolk, UK). In his original version, the prompt included values for a and b (2 and 3 respectively). In the prompt’s more open form (as shown above), students might suggest their own values to substitute into the inequality in order to develop conjectures about the relationship between a and b. To date, students’ questions and observations about the prompt have included:
 What are they trying to find out?
 What values of a and b satisfy the inequality?
 What does the sign mean?
 a could be 2 and b could be 4.
 This can only work when a ≥ 2 and b ≥ 3 with b > a.
 What if a and b are negative numbers? Would the inequalities be reversed?
 What if a and b are fractions?
 Is it possible to change the order of the terms and expressions and find the same values of a and b that satisfy all permutations?
 What if we changed the operations?
 Could we use square roots?
 This one must work all the time: ^{a}/_{b} < a  b < a + b < ab.
 Is it possible for a^{2}, a + b, ab, b^{2 }to be equal?
Inquiries have often led students into changing the order of the terms in the inequality. Other pathways include substituting decimal and negative numbers into the inequality, and using more complex expressions. Five examples of the many conjectures about the relationship between a and b that have arisen in classroom inquiry are:  "If a and b are consecutive whole numbers, then the inequality will never work." By making a = n, a year 9 student showed how the four parts of the inequality become n^{2} < 2n + 1 < n^{2 }+ n < n^{2 }+ 2n + 1. The conjecture turns out to be correct, except when n = 2.
 "The values of a and b must be in the same times table to make this correct." A counterexample (like those in the table below) show this conjecture to be false.
 "As long as a is smaller than b, the prompt will work." This can be shown to be false with a counterexample, such as a = 3, b = 5.
 "If a = 1, then the inequality will never work." This is true because when a = 1, ab < a + b.
 "The inequality will not work if both numbers are negative." This is also true because if a and b are negative, a^{2} > a + b.
One notable result came from a group of students who decided to find the lowest value of b when a is set as a particular whole number.
Value of a  Lowest value b can take given a  First difference  2  3   3  7  4  4  13  6  5  21  8  6  31  10  7  43  12  n  (n  1)^{2} + (n  1) + 1 = n^{2 } n + 1  
The students explained their results by focusing on a^{2} and a + b. In order for a + b > a^{2}, b = a^{2 } a + 1.
Mark Greenaway runs his own website, which is highly recommended for its comprehensive coverage of ideas for the mathematics classroom. Mark can be followed on twitter @suffolkmaths.
Resources
An alternative prompt In Mathematics Teaching 223 (July 2011), Geoff Tennant describes using this prompt (right) with a class on his subject knowledge enhancement course (preteacher training) at Reading University. In order to keep the arithmetic simple, the students decided to substitute perfect squares into the inequality. This led to the following results: a  b  √ab  ½(a+b)  1  4  2  2.5  4  9  6  6.5  9  16  12  12.5  Geoff was taken aback by the results. A quick check of a = 1 and b = 9 led to the conjecture: If a and b are consecutive perfect squares, then ½(a+b)  √ab = ½ The class went on to prove the result by putting a = n^{2} and b = (n+1)^{2}.
 Students' rich questions Amanda Kirby used the substitution prompt for an observation lesson with her year 10 foundation GCSE class. She reports that the students were "fabulous at asking questions and questioning each other" (see their initial comments and questions below). The class quickly realised that the prompt is never true if a and b are negative numbers because, in that case, a^{2} > a + b. As the inquiry progressed, students created their own inequality: a  b^{2 }< a^{2}  b. They wondered if a has to be greater than b for the inequality to be true. Overall, Amanda was pleased with the way the classroom was 'buzzing' during the inquiry and with students' willingness to challenge each other's reasoning.
Amanda Kirby teaches maths and is the Numeracy Across the Curriculum Coordinator at St Clement Danes School, Hertfordshire (UK). Inquiry maths prompts, Amanda says, have proved successful in increasing personalisation, differentiation and students' independence. You can follow Amanda on twitter @mathsteach2000.
Classroom inquiry This prompt was devised for a year 9 class with high prior attainment. The class had carried out mathematical inquiries before and were beginning to inquire independently. Students came up with a wide range of questions and observations that the teacher or, in the case of the example with a = 2 and b = 5, a student wrote on the board (see below). Another student explained that as b^{2}/a is greater than a^{2}/b, b must be greater than a. The class was then given the opportunity to select a regulatory card to decide how the inquiry should proceed. Most students chose to Try to find more examples by which they meant to substitute more values for a and b into the inequality. A minority, however, decided to determine the truth of the inequality. One table of four students wanted to Use a worksheet  a card they had selected on previous occasions  even though it was not included in the set. The teacher gave those students the structured inquiry sheet (see Resources left). As they explored, students began to appreciate that it was difficult to find values for a and b that satisfy the central inequality. One pair of girls proved that the middle two expressions should be reversed by creating two equivalent fractions with the denominator ab and using the inequality b > a to show that ab + b^{2} > a^{2} + b^{2}. Another student proved the prompt was false by showing the middle inequality contradicted the condition b > a. He presented his proof (above) to the class at the end of the lesson. If the middle inequality is correct, he argued, then b < a, which we know to be false.
