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Well done to Amrit, Adithya, Daven and Sergio who all sent in solutions to this problem. Here are the first four inequalities:

$$ 10 < a+ b- c - d < 20 $$
$$ 0 < a- c < 10 $$
$$ -10 < a - c + d - b < 10 $$
$$ 0 < abcd < 625 $$
Aditha's solution explains how to get each of them, you can read the pdf

Amrit explained how to work out the fifth inequality:

For the last inequality, we need to prove the AM-GM inequality
$\frac{a+c}{2}>\sqrt{ac}$
$a+c>2\sqrt{ac}$
$a^2+2ac+c^2>4ac$
$a^2-2ac+c^2>0$
$a-c)^2>0$
A number squared is always greater than 0 unless the number is 0 or in this case if a=c
Plugging in a=c into the last inequality, we have
$\frac{|a|+|c|}{2}-\sqrt|ac|>0$
Looking at the AM-GM inequality, we want a and c to be as far apart as
possible. So |a| has to be 5 and |c| has to be 0 or vice versa.
Applying this, we have $\frac{|a|+|c|}{2}-\sqrt|ac|<\frac{5}{2}$