Question #af811

1 Answer

Suppose that
[a] x>0
[b] 1/x +x=2
are two condition that hold and suppose that this implies not x^2+1/x^2=1/x+x. Then either x^2+1/x^2<1/x+x or x^2+1/x^2>1/x+x has to hold.

Suppose it's x^2+1/x^2<1/x+x. Then using [b] on the right term we get x^2+1/x^2<2, both terms of which can be multiplied by x^2 (x ne 0 thanks to [a]) to get x^4-2x^2+1<0. We recognize the structure of a square of a polynomial, so the inequality can be written as (x^2-1)^2<0. This inequality is never true. So
[1] x^2+1/x^2>1/x+x
has to be valid. We prove by contradiction that it's not.

We can multiply [1] by x^2 ne 0 both terms ([a] guarantees x ne 0) and get x^4-x^3-x+1>0. We now factor the quartic polynomial and obtain (x-1)^2(x^2+x+1)>0 which leads to
[2] x^2+x+1>0
because (x-1)^2=0 iff x=1 and the condition [1] doesn't hold when x=1.

[2] can be written as x+1<-x^2 and thanks to [a] we can divide both sides by x to get 1+1/x<-x. Let's substitute [b] in this inequality and get 2<-x, which is x<-2. This contradicts [a] and the proof is done.

NOTE: I think there should be a shorter and better way of proving the proposition by contradiction. I didn't find it. Let's see if anyone comes up with an idea.