Prove or give a counterexample: for all x > 0 we have x2+1< (x+1)2[tex]\leq[/tex] 2(x2+1)
The Attempt at a Solution
I used my calculator to do the graph of all 3 functions and saw that the statement was always true (at least for x>0). So I couldn't see another proof method that would work so I just went ahead with the direct proof.
Kept rewriting the functions and narrowing down until I got to this below.
-(1/x) < 2 - (1/x) [tex]\leq[/tex] x
My intuition says that if x>0 that when x gets smaller and smaller, the first two terms get more and more negative. As x gets bigger and bigger, the term on the right becomes larger while the other only get slightly larger. However, I didn't see that as a proper "proof."
If it is any help, my class has learned these proof methods so far: Direct Proof, Counterexample, Contrapositive, Contradiction, and proof by cases.
Edit: Found something, I managed to do some algebra and rearranged it to the form of -2x<0<(x-1)(x-1) and pretend that second inequality includes "or equal to." There is a square on the right which will always be positive or 0 and in the problem it says x>0 so that first term will always be less than 0. Sweet.
If something is wrong in my logic please tell me.