Consider
b^(b-a) + a^(b-a) > 0
=> Multiplying both sides by b^(b-a) -a^(b-a) we get
[b^(b-a) + a^(b-a) ][b^(b-a) -a^(b-a)] > 0
=> Difference of two squares
b^(2(b-a)) - a^(2(b-a)) > 0
b^(2(b-a)) > a^(2(b-a)
=>take square root
b^(b-a) > a^(b-a)
=>change negative power into fraction
b^b / b^a > a^b / a^a
=> Multiply both sides by by a^a . b^a
a^a . b^b > a^b . b^a
i think this works out nicely
- using reverse method ![Smile :) :)](data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7)
b^(b-a) + a^(b-a) > 0
=> Multiplying both sides by b^(b-a) -a^(b-a) we get
[b^(b-a) + a^(b-a) ][b^(b-a) -a^(b-a)] > 0
=> Difference of two squares
b^(2(b-a)) - a^(2(b-a)) > 0
b^(2(b-a)) > a^(2(b-a)
=>take square root
b^(b-a) > a^(b-a)
=>change negative power into fraction
b^b / b^a > a^b / a^a
=> Multiply both sides by by a^a . b^a
a^a . b^b > a^b . b^a
i think this works out nicely