How to easily prove $x+\frac{1}{x} \ge 2 \quad ∀x\in ℝ^+$ [duplicate]

This question already has an answer here:

When I tried to solve some certain math problem (an inequation) for pivate exercise purposes, I had to prove that $x+\frac{1}{x} \ge 2 \quad ∀x\in ℝ^+$, I solved it with tools from differential calculus (prooving that there is a local minimum at $(1,2)$ etc), because this was my only concept. But I guess one can prove this in a much simpler way, but I strangely do not get it — So: How can one prove this the most effective way?

marked as duplicate by user14972, user61527, qwr, Davide Giraudo, user88595Jun 7 '14 at 20:13

• Correct, I am sorry… – Lukas Juhrich Jun 7 '14 at 18:43
• AM-GM: $\frac{x+\frac{1}{x}}{2}\ge\sqrt{x\cdot\frac{1}{x}}$ – derivative Jun 7 '14 at 18:46

Hint

Deduce the desired inequality from $$(x-1)^2\ge0$$

• Seconds apart. :) +1! – Alex Wertheim Jun 7 '14 at 18:10
• Aaaahhhh, I knew the answer would be too simple ;) Must wait a few minutes to accept, though :) – Lukas Juhrich Jun 7 '14 at 18:15

Hint: since $x$ is positive, multiply both sides by $x$. Then subtract $2x$ from both sides. Now factor to find... and now just reverse your steps :)

$$x+ \frac{1}{x} \geq 2 \iff x^2+1 \geq 2x \iff x^2-2x+1 \geq \iff (x-1)^2 \geq 0$$

$$\text{The latter is true since z^2=0 \iff z=0}$$.

$$0 \leq \left(\sqrt{x}-\frac{1}{\sqrt{x}}\right)^2$$