only display df lines that have more fs usage than 80%

Assuming you don't have device names containing spaces (which are a pain when it comes to parsing the output of df):

df -P | awk '0+$5 >= 80 {print}'

Adapt the field number if you want to use your implementation's df output format rather than the POSIX format.

Without the 0+, the comparison would be lexical (9% would then be greater than 80). By using the + binary arithmetic operator, we force $5 to be converted to a number (so 9% becomes 9) and the comparison to be numerical. Using the + unary operator (as in awk '+$5 >= 80') works in some awk implementations but not in traditional ones (the ones written by A, W and K) where that operator is just ignored.