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

Question

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?

Correct, I am sorry… – Lukas Juhrich Jun 07 '14 at 18:43 — Lukas Juhrich, Jun 07 '14 at 18:43
AM-GM: $\frac{x+\frac{1}{x}}{2}\ge\sqrt{x\cdot\frac{1}{x}}$ – derivative Jun 07 '14 at 18:46 — derivative, Jun 07 '14 at 18:46

score 6 · Accepted Answer · answered Jun 07 '14 at 18:10

6

Hint

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

answered Jun 07 '14 at 18:10

Seconds apart. :) +1! – Alex Wertheim Jun 07 '14 at 18:10
Aaaahhhh, I knew the answer would be too simple ;) Must wait a few minutes to accept, though :) – Lukas Juhrich Jun 07 '14 at 18:15

score 3 · Answer 2 · answered Jun 07 '14 at 18:10

3

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 :)

answered Jun 07 '14 at 18:10

Alex Wertheim

20,278

Mr.Fry · Answer 3 · 2014-06-07T18:51:21.190

3

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

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

edited Jun 07 '14 at 18:51

answered Jun 07 '14 at 18:39

Mr.Fry

5,003
3
19
28

score 2 · Answer 4 · answered Jun 07 '14 at 18:11

2

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

answered Jun 07 '14 at 18:11

Rene Schipperus

39,526

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

4 Answers4