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?
4 Answers
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Hint
Deduce the desired inequality from $$(x-1)^2\ge0$$
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$\begingroup$ Aaaahhhh, I knew the answer would be too simple ;) Must wait a few minutes to accept, though :) $\endgroup$ Jun 7, 2014 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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$$x+ \frac{1}{x} \geq 2 \iff x^2+1 \geq 2x \iff x^2-2x+1 \geq \iff (x-1)^2 \geq 0$$
$$\text{The latter is true since $z^2=0 \iff z=0$} $$.