Search All of the Math Forum:

Views expressed in these public forums are not endorsed by NCTM or The Math Forum.

Notice: We are no longer accepting new posts, but the forums will continue to be readable.

Topic: Second to last algebra question
Replies: 2   Last Post: May 1, 2005 9:12 PM

 Messages: [ Previous | Next ]
 Elana Posts: 22 Registered: 12/6/04
Second to last algebra question
Posted: Apr 30, 2005 2:47 PM

Hello, all. Digging though my last home assignment here, and stuck on
2 problems once again. Here's the first one:

Prove that a^2 + b^2 + 1 > a + b + ab.

I'm considering two approaches. First one is to consider all cases:

When a and b are < 0, the statement is obviously true, since left side
will always be >0 and right side will always be < 0
When a and b are equal and are equal to 0, the left side is greater,
since it has a constant.
When a and b are equal and are equal to 1, both sides are the same, and
thus the equation still holds.
My only troubles are when a and b are between 0 and 1, and when they
are greater than one. Cannot find a proof without algebraic
manipulation.

Second approach: algebraically prove the ineqality.
I rewrote the inequality as
a^2 + b^2 + 2ab > a + b + 3ab -1, leading to (a + b)^2 > a + b + 3ab
-1. This has not lead me to any meaningful solution.
I also tried writing as a^2 - a + b^2 - b > ab -1, leading to a(a-1) +
b(b-1) > ab - 1. Stuck again.