154710

1 For the inequality involving a-c
2 We need to pair the largest possible value of a and the smallest possible
3 value of c and vice versa
4 a<5 A
5 c>-5 B
6 B*-1:-c<5 B'
7 A+B': a-c<10
8 Now for the smallest value of a and the largest value of c
9 a>-5 A
10 c<5 B
11 B*-1:-c>-5 B'
12 A+B': a-c>-10
13 Collating these results, we have
14 -10<a-c<10 A
15 The same treatment of d and b will give the result
16 -10<d-b<10 B
17 A+B: -20<a-c+d-b<20
18 The largest value of abcd will be obtained by multiplying whole numbers
19 We want to multiply the largest absolute values of the variables together
20 since we are multiplying four variables, meaning the result will be positive
21 The maximum value of a,b,c and d is 5, so abcd<5^4 or abcd<625
22 This answer can also be obtained by realising that the minimum values
23 for all the variables is -5 and that taking (-5)^4 is equivalent to taking (5^4)
21 The smallest value of abcd is obtained if an odd number of
22 variables (in this case 1 or 3 variables) is negative, meaning that abcd is
24 negative. We want these values to have the largest absolute value possible
25 which is 5 but they have to be very close to negative 5. Therefore
26 -5^4<abcd<5^4
27 For the last inequality, we need to prove the AM-GM inequality
28 (a+c)/2>sqrt(ac)
29 a+c>2sqrt(ac)
30 (a^2)+2ac+(c^2)>4ac
31 (a^2)-2ac+(c^2)>0
32 (a-c)^2>0
33 A number squared is always greater than 0 unless the number is 0
34 or in this case if a=c
35 plugging in a=c into the last inequality, we have
36 ((abs(a)+abs(c))/2)-sqrt(abs(ac))>0
37 Looking at the AM-GM inequality, we want a and c to be as far apart as
38 possible. That is we want abs(abs(a)-abs(c)) to be as large as possible.
39 In the last inequality, for this to be possible, a has to be 5 or -5 and c has
40 to be 0 and vice versa. Applying this, we have
41 0<((abs(a)+abs(c))/2)-sqrt(abs(ac))<5/2