For $x>0$, $x + \frac1x \ge 2$ and equality holds if and only if $x=1$ [duplicate]

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Prove that for $x>0$, $x + \frac1x \ge 2$ and equality holds if and only if $x=1$.

I have proven that $x+ \frac1x \ge 2$ by re-writing it as $x^2 -2x +1 \ge0$ and factoring to $(x-1)^2\ge0$ which is true because you cannot square something and it be negative.

Now I am stuck on the part where I have to prove equality to hold if and only if $x=1$. Any suggestions?

marked as duplicate by Martin Sleziak, user85798, Yiyuan Lee, Chris Janjigian, user63181 Mar 22 '14 at 12:08

• Your question has nothing to do with "ordered fields" but with calculus and/or algebra...and I think you're suppsoed to prove it is an equality iff $\;x=1\;$ , not that "it is true" – DonAntonio Oct 16 '13 at 17:08
• Inspired by one of lab's answers sometime ago: apply A.M.-G.M. to $\left(x,\dfrac 1x\right)$ to solve the problem at once. – Git Gud Oct 16 '13 at 17:10
• We have not gotten into Calculus in my proofs class yet and this is in the "ordered fields" section of the text so I assumed it was an ordered fields question. Yes I meant I have to prove equality holds if and only if x=1. – user2659030 Oct 16 '13 at 17:18
• What is the only value of $x$ satisfying $(x - 1)^2 = 0$? It's the repeated root of the quadratic: $x = 1$. – Namaste Oct 16 '13 at 17:20

You want to show that $x+\frac{1}{x}=2$ if and only if $x=1$.
Continuing what you did, we have $(x-1)^2=0$ if and only if $x-1=0$ if and only if $x=1$.
$$\left(\sqrt{x} - \frac{1}{\sqrt{x}}\right) ^ 2 \ge 0$$ $$x - 2 + \frac{1}{x} \ge 0$$ $$x + \frac{1}{x} \ge 2$$