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.
