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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?

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marked as duplicate by user14972, user61527, qwr, Davide Giraudo, user88595 Jun 7 '14 at 20:13

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

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

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

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

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

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

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