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 $\frac{|a|+|c|}{2}-\sqrt|ac|<\frac{5}{2}$

$$ 10 < a+ b- c - d < 20 $$

$$ 0 < a- c < 10 $$

$$ -10 < a - c + d - b < 10 $$

$$ 0 < abcd < 625 $$

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 $\frac{|a|+|c|}{2}-\sqrt|ac|<\frac{5}{2}$