1.
Prove that
holds for all real numbers a, b and c
(>= means greater than or equal to)
2.
Prove that if real numbers a, b and c satisfy
then each of a, b, and c is positive
Thanks for the help guys ^_^
I believe both can be solved by contradiction.
1.
Assume the opposite is true:
for all integers.
or for simplicity
we can instantly see by inspection that any integer does not work since is false.
2.
From and assuming that inequality is correct you can instantly get .
If you start off by assuming one or more variable is less than zero and also assume this set of equations work for all positive or negative integers (depending upon which variable you assume is less than or greater than zero) you will run into contradictions
EDIT: I didn't see the first two replies prior to posting this.. you guys are too quick for me
Here's the second. From the third equation so . Also due to the sign of the third equation either all are positive or two of the three are negative. We will show only the first case is true. Assume that and are negative. Let and so and .
Thus
and
.
Multiplying the inequalities gives
which leads to a contradiction since .
Too slow but with details